Question

Sin^2x+sin^2(60+x)+sin^2(60-x)

Answer 1

Step-by-step explanation:

Given:-

Sin^2x+Sin^2(60°+x)+Sin^2(60°-x)

To find:-

Find the value of Sin^2x+Sin^2(60°+x)+Sin^2(60°-x)

Solution:-

Given expression is

Sin^2x+Sin^2(60°+x)+Sin^2(60°-x)

It can be written as

Sin^2 x + [Sin(60°+x)]^2+[Sin(60°-x)]^2

We know that

Sin (A+B)= Sin A Cos B + Cos A Sin B

Sin (A-B)= Sin A Cos B - Cos A Sin B

Where A = 60° and B= x

Sin (60°+x)

=> Sin 60° Cos x + Cos 60° Sin x

=> (√3/2) Cos x + (1/2) Sin x

=> (√3 Cos x + Sin x)/2

[Sin (60+x)]^2

=> [(√3 Cos x + Sin x)/2]^2

[(√3Cosx)^2+2(√3Cosx)(Sinx)+(Sinx)^2]/4

=> [3Cos^2 x+2√3 SinxCosx+Sin^2 x ]/4

and

Sin (60°-x)

=> Sin 60° Cos x - Cos 60° Sin x

=> (√3/2) Cos x - (1/2) Sin x

=> (√3 Cos x - Sin x)/2

[Sin (60-x)]^2

=> [(√3 Cos x - Sin x)/2]^2

[(√3Cosx)^2-2(√3Cosx)(Sinx)+(Sinx)^2]/4

[3Cos^2 x-2√3SinxCosx+Sin^2 x ]/4

Now ,

Sin^2(60+x) + Sin^2 (60-x)

=>(3Cos^2 x+2√3Sinx Cos x + Sin^2 x)/4 +

( 3 Cos^2 x -2√3 Sin x Cos x + Sin^2 x )/4

=>( 6 Cos^2 x + 2 Sin^2 x)/4

=>(3Cos^2 x + Sin^2 x)/2

Now ,

Sin^2x+Sin^2(60+x)+Sin^2(60-x)

=> Sin^2 x + (3Cos^2 x + Sin^2 x)/2

=> (2 Sin^2 x + 3 Cos^2 x+Sin^2 x)/2

=>( 3 Sin^2 x+ 3 Cos^2 x )/2

We know that

Sin^2 A + Cos^2 A = 1

=> [3(1)]/2

=> 3/2

Answer:-

The value of Sin^2x+Sin^2(60°+x)+Sin^2(60°-x) is

3/2

Used formulae:-

• Sin (A+B)= Sin A Cos B + Cos A Sin B

• Sin (A-B)= Sin A Cos B - Cos A Sin B

• Sin^2 A + Cos^2 A = 1

• Sin 60°=√3/2

• Cos 60°=1/2