Mathematics Trigonometric Functions of Sum and Difference of Two Angles , Sum Of Two Functions and Product Of Two Functions
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\color{red} {⧫} Trigonometric Functions of Sum and Difference of Two Angles
\color{red} {⧫} Sum Of Two Functions
\color{red} {⧫} Product Of Two Functions

### Trigonometric Functions of Sum and Difference of Two Angles

\color{blue} ✍️ In this Section, we shall derive expressions for trigonometric functions of the sum and difference of two numbers (angles) and related expressions.

The basic results in this connection are called trigonometric identities. We have seen that
1. \ \ color{red}(sin (– x) = – sin x)
2. \ \ color{red}( cos (– x) = cos x)

We shall now prove some more results:

3. \ \ color{red}( cos (x + y) = cos x cos y – sin x sin y)

\color{blue} ✍️ Consider the unit circle with centre at the origin. Let x be the angle color{red}(P_4OP_1) and y be the angle color{red}(P_1OP_2). Then color{red}((x + y)) is the angle color{red}(P_4OP_2). Also let color{red}((– y)) be the angle color{red}(P_4OP_3).

\color{blue} ✍️ Therefore, P_1, P_2, P_3 and P_4 will have the coordinates color{red}(P_1 \ \(cos x, sin x), color{greeen}[ P_2 \ \ [cos (x + y), sin (x + y)]],color{purple}[ P_3 \ \ [cos (– y), sin (– y)]] and color{red}(P_4 \ \ (1, 0)) (Fig 3.14).

Consider the triangles color{red}(P_1OP_3) and color{red}(P_2OP_4). They are congruent Therefore,

color{red}(1P_1P_3) and color{red}(1P_2P_4) are equal. By using distance formula, we get

color{red}(1P_1P_3^2 = [cos x – cos (– y)]^2 + [sin x – sin(–y)^2]

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \= (cos x – cos y)^2 + (sin x + sin y)^2

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \= cos^2 + cos^2 y – 2 cos x cos y + sin^2 x + sin^2 y + 2sin x sin y

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \= 2 – 2 (cos x cos y – sin x sin y)

Also, color{red}(P_2P_4^2 = [1 – cos (x + y)]^2 + [0 – sin (x + y)]^2)

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \= 1 – 2cos (x + y) + cos^2 (x + y) + sin^2 (x + y)

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \= 2 – 2 cos (x + y)

Since color{red}(P_1P_3 = P_2P_4,) we have color{red}(P_1P_3^2 = P_2P_4^2).

Therefore, 2 –2 (cos x cos y – sin x sin y) = 2 – 2 cos (x + y).

Hence color{red}(cos (x + y) = cos x cos y – sin x sin y)

### 4.  cos (x – y) = cos x cos y + sin x sin y

Replacing y by – y in identity cos (x + y) = cos x cos y – sin x sin y, we get

cos (x + (– y)) = cos x cos (– y) – sin x sin (– y)

or color{red}(cos (x – y) = cos x cos y + sin x sin y)

### 5 . cos (pi/2 -x) = sin x

If we replace x by color{red}(pi/2) and y by x in Identity (4), we get

color{red}(cos (pi/2-x) = cos (pi/2) cos x + sin ( pi/2) sin x = sin x)

### 6. sin (pi/2 -x) = cos x

Using the Identity 5, we have

color{red}(sin (pi/2 -x) = cos [ pi/2 - (pi/2- x) ] = cos x)

### 7. sin (x + y) = sin x cos y + cos x sin y We know that

color{blue}(sin (x+ y) = cos (pi/2 - (x+y) ) = cos ( (pi/2 -x ) - y ))

color{}(= cos (pi/2-x) cos y +sin (pi/2-x) sin y)

color{red}(= sin x cos y + cos x sin y)

### 8 sin (x – y) = sin x cos y – cos x sin y

If we replace y by –y, in the Identity 7, we get the result .

color(red)(sin (x – y) = sin x cos y – cos x sin y)
Q 3136601572

Find the value of sin 15°.

Solution:

We have
sin 15° = sin (45° – 30°)
= sin 45° cos 30° – cos 45° sin 30°

=1/sqrt2 xx sqrt3/2 - 1/2 xx 1/2 = ((sqrt3)-1)/2sqrt2
Q 3186701677

Prove that

(sin (x+y))/(sin (x-y))= (tan x+tan y)/(tan x-tan y)

Solution:

We have

L.H.S. (sin (x+y))/(sin (x-y)) = (sin x cos y + cos xsin y)/(sin x cos y - cos x sin y )

Dividing the numerator and denominator by cos x cos y, we get

(sin (x+y))/(sin (x-y))= (tan x+tan y)/(tan x-tan y)

### 9. By taking suitable values of x and y in the identities 3, 4, 7 and 8, we get the following results:

color{red}(cos ( pi/2 +x) = - sin x)

color{red}(sin (pi/2 +x) = cos x)

color{red}(cos (pi-x) = - cos x)

color{red}(sin (pi-x) = sin x)

color{red}(cos (π + x) = – cos x)

color{red}(sin (π + x) = – sin x)

color{red}(cos (2π – x) = cos x)

color{red}(sin (2π – x) = – sin x)

Similar results for tan x, cot x, sec x and cosec x can be obtianed from the results of sin x and cos x.
Q 3106501478

Prove that

3sin (pi/6) sec(pi/3) -4 sin (5pi/6) cot pi/4 = 1

Solution:

We have

L.H.S. = 3sin (pi/6) sec( pi/6) - 4 sin ( 5pi/6) cot pi/4

= 3 ×1/2 xx 2 -4 sin (pi - pi/6 ) xx 1 =3 -4 sin  pi/6

= 3 – 4 × 1/2 = 1 = R.H.S

### 10. If none of the angles x, y and color{red}((x + y)) is an odd multiple of color{red}(pi/2) , then  tan (x+y) = (tan x + tan y)/(1- tan x tan y)

Since none of the color{red}(x, y) and color{red}((x + y)) is an odd multiple of color{red}(π/2) , it follows that cos x, cos y and cos (x + y) are non-zero. Now

color{green}( tan (x + y) = (sin (x+y ) )/( cos (x+y) ) = (sin x cos y + cos x sin y )/( cos x cos y - sin x sin y ))

Dividing numerator and denominator by cos x cos y, we have

color{red}(tan (x+y) ) = ( (sin x cos y )/( cos x cos y ) + ( cos x sin y)/(cos x cos y) )/( (cos x cos y)/( cos x cos y ) - ( sin x sin y )/( cos x cos y) )

 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \color{red}(= (tan x + tan y )/(1- tan x tan y ))

Q 3106701678

Show that
tan 3 x tan 2 x tan x = tan 3x – tan 2 x – tan x

Solution:

We know that 3x = 2x + x

Therefore, tan 3x = tan (2x + x)

or tan3x = (tan 2x+tanx)/(1-tan 2sx tan x)

or tan 3x – tan 3x tan 2x tan x = tan 2x + tan x
or tan 3x – tan 2x – tan x = tan 3x tan 2x tan x
or tan 3x tan 2x tan x = tan 3x – tan 2x – tan x.

### tan ( x – y) = (tan x - tan y )/( 1+ tan x tan y)

If we replace y by – y in Identity 10, we get

color{blue}(tan (x – y) = tan [x + (– y)])

 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ color{red}(= (tan x+ tan (-y) )/( 1- tan x tan (-y) ) = (tan x - tan y)/(1+ tan x tan y ))

Q 3126701671

Find the value of tan ((13pi)/12)

Solution:

We have
tan ((13pi)/12) = tan (pi+pi/12) = tan (pi/12) = tan (pi/4 - pi/6)

(tan (pi/4) - tan (pi/6))/(1+ tan (pi/4) tan (pi/6)) = (1-1/sqrt3)/(1+1/sqrt3) = (sqrt(3)-1)/(sqrt3+1) = 2 - sqrt3

### If none of the angles x, y and (x + y) is a multiple of π, then

color{red}(cot (x+ y ) = (cot x cot y -1 )/( cot y + cot x))

Since, none of the x, y and (x + y) is multiple of π, we find that sin x sin y and sin(x + y) are non-zero. Now,

color{green}(cot (x+ y) = (cos (x+ y) )/(sin (x+ y) ) = (cos x cos y – sin x sin y)/( sin x cos y + cos x sin y))

Dividing numerator and denominator by sin x sin y, we have

color{red}( cot (x+ y) = (cot x cot y -1)/( cot y + cot x))

### cot (x – y)= (cot x cot y + 1)/(cot y - cot x)

If we replace y by –y in identity 12, we get the result

color(red)(cot (x – y)= (cot x cot y + 1)/(cot y - cot x))

### cos 2x = cos^2 x – sin^2 x = 2 cos^2 x – 1 = 1 – 2 sin^2 x = (1- tan^2 x)/(1+ tan^2 x)

\color{blue} ✍️ We know that

color{green}(cos (x + y) = cos x cos y – sin x sin y)

\color{blue} ✍️ Replacing y by x, we get

color{red}(cos 2x )= cos^2x – sin^2 x = 2 cos^2 x – 1

 \ \ \ \ \ \ \ \ = cos^2 x – (1 – cos^2 x) =color{red}( 2 cos^2x – 1)

\color{blue} ✍️ Again, color{green}(cos 2x = cos^2 x – sin^2 x)

 \ \ \ \ \ \ \ \ \ \ = 1 – sin^2 x – sin^2 x = color{red}( 1 – 2 sin^2 x).

\color{blue} ✍️ We have color{green}(cos 2x = cos^2 x – sin^2 x = (cos^2 x - sin^2 x)/( cos^2 x + sin^2 x))

Dividing each term by color{orange}(cos^2 x,) we get

color{red}(cos 2x = (1- tan^2 x)/(1+ tan^2 x))

### sin 2x = 2 sinx cos x = (2 tan x)/( 1+ tan^2 x)

\color{blue} ✍️ We have

color{red}(sin (x + y) = sin x cos y + cos x sin y)

\color{blue} ✍️ Replacing y by x,

we get color{red}(sin 2x = 2 sin x cos x.)

\color{blue} ✍️ Again color{green}(sin 2x = (2 sin x cos x)/( cos^2 x + sin^2 x))

Dividing each term by color{red}(cos^2 x), we get

color{red}(sin 2x = (2 tan x )/( 1+ tan^2 x))

### tan 2x = (2 tan x)/(1- tan^2 x)

\color{blue} ✍️ We know that

color{green}(tan (x+y) = (tan x + tan y )/( 1- tan x tan y))

Replacing y by x ,

we get color{red}(tan 2x = (2 tan x)/(1 - tan^2 x))

### sin 3x = 3 sin x – 4 sin^3 x

\color{blue} ✍️ We have,

color{green}(sin 3x = sin (2x + x))

 \ \ \ \ \ \ \ \ \ \ \= sin 2x cos x + cos 2x sin x

\ \ \ \ \ \ \ \ \ \ \= 2 sin x cos x cos x + (1 – 2sin^2 x) sin x

\ \ \ \ \ \ \ \ \ \ \= 2 sin x (1 – sin^2 x) + sin x – 2 sin^3 x

\ \ \ \ \ \ \ \ \ \ \= 2 sin x – 2 sin^3 x + sin x – 2 sin^3 x

\ \ \ \ \ \ \ \ \ \ \ color{red}(= 3 sin x – 4 sin^3 x)

### cos 3x= 4 cos^3 x – 3 cos x

\color{blue} ✍️ We have,

color{green}(cos 3x = cos (2x +x))

\ \ \ \ \ \ \ \ \ \ \= cos 2x cos x – sin 2x sin x

\ \ \ \ \ \ \ \ \ \ \= (2cos^2 x – 1) cos x – 2sin x cos x sin x

\ \ \ \ \ \ \ \ \ \ \= (2cos^2 x – 1) cos x – 2cos x (1 – cos^2 x)

\ \ \ \ \ \ \ \ \ \ \= 2cos^3 x – cos x – 2cos x + 2 cos^3 x

\ \ \ \ \ \ \ \ \ \ \ color{red}(= 4 cos^3 x – 3cos x).

### tan 3x = (3 tan x - tan^3 x)/( 1-3 tan^2 x)

\color{blue} ✍️ We have

color{red}(tan3x =tan (2x + x))

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ color{green}(= (tan 2x +tanx)/( 1- tan 2x tan x))

 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = ( (2 tan x)/( 1- tan^2 x) + tan x)/( 1- ( 2 tan x * tan x )/(1- tan^2 x) )

 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \=(2 tan x + tan x - tan^3 x)/(1- tan^2 x -2 tan^2 x)

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ color{red}( = (3 tan x - tan^3 x)/( 1- 3 tan^2 x))

### Sum Of Two Functions

(i) color{red}(cos x + cos y = 2 cos( (x+y)/2) cos ((x-y)/2))

(ii) color{red}(cos x – cos y = -2 sin ((x+y)/2) sin ((x-y)/2))

(iii) color{red}(sin x + sin y = 2 sin ((x+y)/2 ) cos ((x-y)/2))

(iv) color{red}(sin x – sin y = 2 cos ((x+y)/2) sin ((x-y)/2))

\color{blue} ✍️ We know that

color{green}(cos (x + y) = cos x cos y – sin x sin y) .................................................... (1)

and color{green}(cos (x – y) = cos x cos y + sin x sin y ).............................................. (2)

color{blue}("Adding and subtracting (1) and (2)"), we get

color{orange}(cos (x + y) + cos(x – y) = 2 cos x cos y) .................................... (3)

and color{orange}(cos(x + y) – cos (x – y) = – 2 sin x sin y) .....................................................(4)

\color{blue} ✍️ We know that

color{green}(sin (x + y) = sin x cos y + cos x sin y) .................................................. (5)

and sin color{green}((x – y) = sin x cos y – cos x sin y) .............................................. (6)

color{blue}("Adding and subtracting (5) and (6)"), we get

color{orange}(sin (x + y) + sin (x – y) = 2 sin x cos y) ................................................ (7)

color{orange}(sin (x + y) – sin (x – y) = 2cos x sin y) .............................................. (8)

Let color{red}(x + y = θ and x – y = φ.) Therefore

color{red}(x= ( ( θ +φ )/2 )) and color{red}(y = ( (θ -φ )/ 2))

Substituting the values of x and y in (3), (4), (7) and (8), we get

color{red}(cos θ + cos φ = 2 cos ( ( θ +φ)/2 ) cos ((θ -φ)/2 ))

color{red}(cos θ – cos φ = – 2 sin ( ( θ +φ )/2) sin ( ( θ -φ)/2))

color{red}(sin θ + sin φ = 2 sin ( ( θ +φ )/2 ) cos ( ( θ - φ)/2))

color{red}(sin θ – sin φ = 2 cos ( (θ +φ )/2) sin ( ( θ - φ)/2 ))

Since θ and φ can take any real values, we can replace θ by x and φ by y.
Thus, we get

color{red}(cos x + cos y = 2 cos( (x+y)/2) cos ((x-y)/2)) ; \ \ \ \ \ \ \ \ \ \color{red}(cos x – cos y = -2 sin ((x+y)/2) sin ((x-y)/2)) ,

color{red}(sin x + sin y = 2 sin ((x+y)/2 ) cos ((x-y)/2)) ; \ \ \ \ \ \ \ \ \ \ \ color{red}(sin x – sin y = 2 cos ((x+y)/2) sin ((x-y)/2)) .
Q 3116801770

Prove that

cos (pi/4 +x) + cos (pi/4 - x) = sqrt2cos x

Solution:

Using the Identity 20(i), we have

L.H.S cos (pi/4 +x) + cos (pi/4 - x)

= 2 cos ((pi/4 + x + pi/4 - x)/2) cos ((pi/4 + x - (pi/4 - x))/2)

= 2 cos(pi/4 )cos x = 2 xx 1/sqrt2 cos x = sqrt2 cos x = R.H.S
Q 3176801776

Prove that (cos 7x + cos 5x)/(sin 7x - sin 5x) = cot x

Solution:

Using the Identities 20 (i) and 20 (iv), we get

L.H.S (2 cos ((7x+5x)/2)cos((7x-5x)/2))/(2cos ((7x+5x)/2) sin ((7x-5x)/2)) = (cosx)/(sinx) = cot x = R.H.S
Q 3136001872

Prove that = (sin 5x -2 sin 3x + sinx)/(cos5x-cosx) = tan x

Solution:

We have L.H.S = (sin 5x-2 sin 3x+sinx)/(cos 5x - cosx) = (sin5x+sinx-2 sin 3x)/(cos 5x- cos)

= (2sin 3 x cos 2x - 2sin 3x)/(-2sin 3x sin 2x) = - (sin 3 x (cos2x-1))/(sin 3x sin 2x)

= (1-cos2x)/(sin2x) = (2sin^2x)/(2 sin xcos x) = tan x = R.H.S

### Product Of Two Functions

(i) color{red}(2 cos x cos y = cos (x + y) + cos (x – y))

(ii) color{red}(–2 sin x sin y = cos (x + y) – cos (x – y))

(iii) color{red}(2 sin x cos y = sin (x + y) + sin (x – y))

(iv) color{red}(2 cos x sin y = sin (x + y) – sin (x – y)).