Question

sin (45+x) sin (45-x) = 1/2 cos 2x​

Answer 1

Answer:

sin(45+x)·sin(45-x) = (1/2)cos2x

 

Use the sum and difference of two angles identity:

    (sin45cosx+sinxcos45)·(sin45cosx-sinxcos45) = (1/2)cos2x

 

This is of the form (a+b)(a-b) = a2 - b2, so it's:

    sin245cos2x - sin2xcos245 = (1/2)cos2x

 

Use The Pythagorean Identity sin2θ+cos2θ=1, so sin2θ=1-cos2θ.  Also sin245 = cos245 = 1/2

    (1/2)cos2x - (1-cos2x)(1/2) = (1/2)cos2x

    (1/2)(cos2x - 1 + cos2x) = (1/2)cos2x

    (1/2)(2cos2x - 1) = (1/2)cos2x

 

Last identity: cos2x = 2cos2x - 1:

    (1/2)cos2x = (1/2)cos2x

Step-by-step explanation:

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Answer 2

We know that sin(A+B)=sinAcosB+sinBcosA

and sin(A-B)=sinAcosB-sinBcosA

So, sin(45+x)sin(45-x)

=(sin45cosx+sinxcos45)(sin45cosB-sinBcos45)

=(1/√2cosx+1/√2sinx)(1/√2cosx-1/√2sinx)

=1/2cos^2x-sin^2x. {(A+B)(A-B)=A^2-B^2}

=1/2cos2x. {cos^2x-sin^2x=cos2x}