MHB Trigonometric identities transformation

[Drain Brain]

I already did everything that I can to transform the left side member to the right side member but I always get a jumbled terms. Please give me a hand on this problem.

$(2\sin^{2}(\theta)-\cos^{2}(\theta))^{2}-9(2\sin^{2}(\theta)-1)^{2}=(2-3\sin^{2}(\theta))(2+3\sin(\theta))(3\sin(\theta)-2)$

 

[ineedhelpnow]

Hey Drain Brain.

ok so for the LHS you have to use a trig sub $cos^2(x)=1-sin^2(x)$

$(2sin^2(x)-(1-sin^2(x)))^2-9(2sin^2(x)-1)^2$ expand this. multiply it out.

for the RHS just expand what you have

$(2-3sin^2(x))(2+3sin(x))(3sin(x)-2)$ = $-8-27\sin ^4\left(x\right)+30\sin ^2\left(x\right)$

make sure the LHS is equal to what I showed for the RHS (you can work out the intermediate steps to show the expansion)

when working on this type of problems, i don't know if it works for you, but i always like to have a sheet of all the trig identities and everything so i can reference to. you may have to play with them a little bit to solve for certain things but NORMALLY you should try to get the whole thing in terms for one thingy only (i forgot what its called) like do it all in terms of sine or all in terms of cosine and if you have tangent you may want to simplify to sine and cosine. so yeah, i hope that helps. and by the way i just realized i did the whole thing using 'x' instead of theta by mistake (Blush) but just make sure not to write 'x' down when you're doing it.

 

[ineedhelpnow]

oh just to note $cos^2(x)=1-sin^2(x)$ is just $sin^2(x)+cos^2(x)=1$ :)

 

[kaliprasad]

I already did everything that I can to transform the left side member to the right side member but I always get a jumbled terms. Please give me a hand on this problem.

$(2\sin^{2}(\theta)-\cos^{2}(\theta))^{2}-9(2\sin^{2}(\theta)-1)^{2}=(2-3\sin^{2}(\theta))(2+3\sin(\theta))(3\sin(\theta)-2)$

I see that there is no $\cos\theta$ on the RHS so convert $\cos^2\theta$ on LHS to $1-\sin ^2\theta$ and get LHS as
$(3\sin^{2}(\theta)-1))^{2}-9(2\sin^{2}(\theta)-1)^{2}$
which is difference of 2 squares.
now you should be able to proceed

 

[soroban]

Hello, Drain Brain!

$(2\sin^2x-\cos^2x)^2-9(2\sin^2x-1)^2\;=\;(2-3\sin^2x)(2+3\sin x)(3\sin x-2)$

Since $\cos^2x \:=\:1-\sin^2x$, the left side becomes:

$\quad (3\sin^2x-1)^2 - 9(2\sin^2x-1)^2\qquad $difference of squares!

$\;\;=\;\big[(3\sin^2x-1) - 3(2\sin^2x-1)\big]\,\big[(3\sin^2x-1) + 3(2\sin^2x - 1)\big] $

$\;\;=\;\big[3\sin^2x-1-6\sin^2x+3\big]\,\big[3\sin^2x-1 + 6\sin^2x-3\big]$

$\;\;=\;(2-\sin^2x)(9\sin^2x-4)$

$\;\;=\; (2-\sin^2x)(3\sin x - 2)(3\sin x + 2)$