# Help in Calculating a Value - Trigonometry

• September 4th 2008, 06:59 AM
Ruler of Hell
Help in Calculating a Value - Trigonometry
Hi, I had this sum for an exam and I couldn't get it, so I just wanted to know how to solve it:

Given: sinA + sin^2A = 1

Calculate: cos^12A + 3cos^10A + 3cos^8A cos^6A 2cos^4A 2cos^2A - 2

I didn't know how to use the code, so I used ^ for power. Sorry for that.

This is what I did:
sinA + sin^2A = 1
sinA = 1 - sin^2A
But, 1 - sin^2A = cos^2A
Therefore, sinA = cos^2A

...And then I substituted cos^2A as sinA throughout this:
cos^12A + 3cos^10A + 3cos^8A cos^6A 2cos^4A 2cos^2A - 2

Yet, I could not arrive at an answer. All help is greatly appreciated.

Thanks,
RoH.
• September 4th 2008, 07:42 AM
Moo
Hello,
Quote:

Originally Posted by Ruler of Hell
Given: sinA + sin^2A = 1

Calculate: cos^12A + 3cos^10A + 3cos^8A cos^6A 2cos^4A 2cos^2A - 2

Are you sure it is not :
$\cos^{12}A + 3\cos^{10}A + 3\cos^8A + \cos^6A + 2\cos^4A + 2\cos^2A - 2$

Ok, when substituting, you should have :
$\sin^6A+3\sin^5A+3\sin^4A+\sin^3A+2\sin^2A+2\sin A-2$

Develop $(\sin A+\sin^2A)^3=1^3$, use $\sin A+\sin^2A=1$ and see what can be simplified =)
• September 4th 2008, 10:14 AM
Soroban
Hello, Ruler of Hell!

Your game plan is a good one.
And I agree with Moo . . . you left out some plus-signs.

Quote:

Given: . $\sin A + \sin^2\!A \:=\:1 \quad\Rightarrow\quad {\color{blue}\sin A(\sin A + 1) \:=\:1}$

Calculate: . $\cos^{12}\!A + 3\cos^{10}\!A + 3\cos^8\!A \:{\color{red}+}\: \cos^6\!A \:{\color{red}+}\: 2\cos^4\!A \:{\color{red}+}\: 2\cos^2\!A - 2$

This is what I did: . $\sin A + \sin^2\!A \:=\: 1 \quad\Rightarrow\quad \sin A \:= \:1 - \sin^2\!A$

But . $1 - \sin^2\!A \:= \:\cos^2\!A$
Hence, $\sin A \:= \:\cos^2\!A$

Substituting, we get: . $\sin^6\!A + 3\sin^5\!A + 3\sin^4\!A + \sin^3\!A + 2\sin^2\!A + 2\sin A - 2$

Factor: . $\sin^3\!A\left(\sin^3\!A + 3\sin^2\!A + 3\sin A + 1\right) + 2\sin x(\sin x + 1) - 2$

Factor: . $\sin^3\!A(\sin A + 1)^3 + 2\sin x(\sin x + 1) - 2$

We have: . $\underbrace{\bigg[\sin A(\sin A + 1)\bigg]^3}_{\text{This is 1}} + 2\underbrace{\bigg[\sin A(\sin A + 1)\bigg]}_{\text{This is 1}} - 2$

Therefore: . $1^3 + 2\!\cdot\!1 - 2 \;\;=\;\;\boxed{1}$

• September 4th 2008, 11:37 AM
Shyam
$
\sin A + \sin ^2 A = 1 \hfill \\$

$\Rightarrow \sin A = 1 - \sin ^2 A \hfill \\$
$\Rightarrow \sin A = \cos ^2 A \hfill \\
\$

we will use this value in the given expression:
Substitute:
$
= \sin ^6 A + 3\sin ^5 A + 3\sin ^4 A + \sin ^3 A + 2\sin ^2 A + 2\sin A - 2
\$

Factoring:
$
=\sin^3\!A\left(\sin^3\!A + 3\sin^2\!A + 3\sin A + 1\right) + 2\sin A(\sin A + 1) - 2
$

$
=\sin^3\!A(\sin A + 1)^3 + 2\sin A(\sin A + 1) - 2
$

$
=[\sin A(\sin A + 1)]^3 + 2[\sin A(\sin A + 1)] - 2
$

$
=[\sin^2 A+\sin A)]^3 + 2[\sin^2 A+\sin A)] - 2
$

$= 1^3+2(1)-2\$

$=1$