Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

( 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.

Answer 5

thank you so much :)

Answer 6

yw :)