Which of the following are identities? More than 1 answer: a. sin x+ sin5x=tan3x b. sin^6 x+cos^6 x=4-3sin^2 2x c. cos^2 3x-sin^2 3x=cos6x d.(sin x+ cos x)^2 =1+sin 2x

Let's check them all a) If x = pi/4, then sin(x)+sin(5x) = tan(3x) sin(pi/4)+sin(5pi/4) = tan(3pi/4) 0.707-0.707= 0-1 0 = -1 which is false So sin(x)+sin(5x) = tan(3x) is NOT an identity

b) If x = 0, then sin^6(x)+cos^6(x)=4-3sin^2(2x) sin^6(0)+cos^6(0)=4-3sin^2(2*0) (0)^6+(1)^6 = 4 - 3*(0)^2 0+1 = 4 - 3*0 0+1 = 4 - 0 1 = 4 which is false, so sin^6(x)+cos^6(x)=4-3sin^2(2x) is NOT an identity

c) cos^2(3x)-sin^2(3x)=cos(6x) cos^2(3x)-sin^2(3x)=cos(2*3x) cos^2(3x)-sin^2(3x)=cos(3x+3x) cos^2(3x)-sin^2(3x)=cos(3x)cos(3x)-sin(3x)sin(3x) cos^2(3x)-sin^2(3x)=cos^2(3x)-sin^2(3x) So cos^2(3x)-sin^2(3x)=cos(6x) is an identity

( sin(x)+ cos(x) )^2 =1+sin(2x) sin^2(x) + 2sin(x)cos(x)+ cos^2(x) =1+sin(2x) [sin^2(x) + cos^2(x)]+ [2sin(x)cos(x)] =1+sin(2x) 1 + sin(2x) = 1+sin(2x) So ( sin(x)+ cos(x) )^2 =1+sin(2x) is an identity.

thank you so much :)

yw :)

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