To begin with i was barking up the wrong tree and trying to prove it was cos squared instead of sin squared, but later on i tried a different route with this in mind and ended up with the same solution
To begin with i was barking up the wrong tree and trying to prove it was cos squared instead of sin squared, but later on i tried a different route with this in mind and ended up with the same solution
ill have a go cos2(x−45)−21(cos2a−sin2a)=sin2x[br][br]cos2(x−45)−21(cos2a−sin2a)=1−cos2x[br][br]cos2(x−45)−21((sin2x−cos2x)−sin2a)=1−cos2x[br][br]cos2(x−45)−21(((1−cos2x)−cos2x)−sin2a)=1−cos2x[br][br]cos2(x−45)−21(1−2cos2x−sin2x)=1−cos2x[br][br]cos2(x−45)−21(1−2cos2x−sin2x)=1−cos2x[br][br]2cos2(x−45)−1+2cos2x+sin2x=2−2cos2x[br][br]2cos2(x−45)−3+4cos2x+sin2x=0[br][br]2cos2(x−45)−3+4cos2x+2(cosx)(sinx)=0[br][br]
ill have a go cos2(x−45)−21(cos2a−sin2a)=sin2x[br][br]cos2(x−45)−21(cos2a−sin2a)=1−cos2x[br][br]cos2(x−45)−21((sin2x−cos2x)−sin2a)=1−cos2x[br][br]cos2(x−45)−21(((1−cos2x)−cos2x)−sin2a)=1−cos2x[br][br]cos2(x−45)−21(1−2cos2x−sin2x)=1−cos2x[br][br]cos2(x−45)−21(1−2cos2x−sin2x)=1−cos2x[br][br]2cos2(x−45)−1+2cos2x+sin2x=2−2cos2x[br][br]2cos2(x−45)−3+4cos2x+sin2x=0[br][br]2cos2(x−45)−3+4cos2x+2(cosx)(sinx)=0[br][br]
Continue from here
Its an identity? Surely you start with the left hand side to prove the right hand side?
ill have a go cos2(x−45)−21(cos2a−sin2a)=sin2x[br][br]cos2(x−45)−21(cos2a−sin2a)=1−cos2x[br][br]cos2(x−45)−21((sin2x−cos2x)−sin2a)=1−cos2x[br][br]cos2(x−45)−21(((1−cos2x)−cos2x)−sin2a)=1−cos2x[br][br]cos2(x−45)−21(1−2cos2x−sin2x)=1−cos2x[br][br]cos2(x−45)−21(1−2cos2x−sin2x)=1−cos2x[br][br]2cos2(x−45)−1+2cos2x+sin2x=2−2cos2x[br][br]2cos2(x−45)−3+4cos2x+sin2x=0[br][br]2cos2(x−45)−3+4cos2x+2(cosx)(sinx)=0[br][br]