Question

solve this trigonomery​

Answer 1

Answer:

Step-by-step explanation:

Take theta as x for convenience,

So we could rewrite the eqn as,

sinx+sin3x= 1-sin2x

(It would be easy for us if we convert sinx in terms of cosx)

=> sinx(1+sin2x)= cos2x (since, sin2x+cos2x=1)

=> sinx(1+1-cos2x)=cos2x

=> sinx(2-cos2x)=cos2x

(then, square on both sides)

=> sin2x [(2-cos2x)]^2= cos4x

=> (1-cos2x) ( 4+cos4x-4cos2x)= cos4x

=> 4+cos4x-4cos2x-4cos2x-cos6x+4cos4x=cos4x

=> 4-4cos2x-4cos2x-cos6x+4cos4x=1

=> 4-8cos2x-cos6x-4cos4x=0

=> cos6x+4cos4x+8cos2x=4...........

Hence, proved!