# Trigonometric function

• Jun 15th 2013, 05:01 AM
Trefoil2727
Trigonometric function
1. By squaring both sides of the identity sin^2 A + cos^2 A=1, or otherwise, show that sin^4 A + cos^4=(1/4)(3+cos4A). Hence find the possible greatest value and the smallest value of sin^4 θ + cos^4 θ

2. Given that A, B, C are the angles of a triangle, show that tan C= (tanA+tanB)/(tanAtanB-1)
• Jun 15th 2013, 05:49 AM
topsquark
Re: Trigonometric function
Quote:

Originally Posted by Trefoil2727
1. By squaring both sides of the identity sin^2 A + cos^2 A=1, or otherwise, show that sin^4 A + cos^4=(1/4)(3+cos4A). Hence find the possible greatest value and the smallest value of sin^4 θ + cos^4 θ

2. Given that A, B, C are the angles of a triangle, show that tan C= (tanA+tanB)/(tanAtanB-1)

What have you been able to do so far?

-Dan
• Jun 16th 2013, 04:48 AM
Trefoil2727
Re: Trigonometric function
(sin^2 A + cos^2 A)^2 =1
sin^4 A + 2sin^2 A cos^2 A + cos^4 A =1
sin^4 A + cos ^4 A = 1- 2sin^2 A cos^2 A

that's all I can do..

For the second question, I really don't know where to start
• Jun 16th 2013, 05:24 AM
Prove It
Re: Trigonometric function
Quote:

Originally Posted by Trefoil2727
(sin^2 A + cos^2 A)^2 =1
sin^4 A + 2sin^2 A cos^2 A + cos^4 A =1
sin^4 A + cos ^4 A = 1- 2sin^2 A cos^2 A

that's all I can do..

For the second question, I really don't know where to start

\displaystyle \displaystyle \begin{align*} \sin^4{(A)} + \cos^4{(A)} &= 1 - 2\sin^2{(A)}\cos^2{(A)} \\ &= 1 - 2\left[ \sin{(A)}\cos{(A)} \right] ^2 \\ &= 1 - 2 \left[ \frac{1}{2}\sin{(2A)} \right] ^2 \\ &= 1 - 2 \cdot \frac{1}{4} \sin^2{(2A)} \\ &= 1 - \frac{1}{2} \sin^2{(2A)} \\ &= 1 - \frac{1}{2} \left[ \frac{1}{2} - \frac{1}{2}\cos{(4A)} \right] \\ &= 1 - \frac{1}{4} + \frac{1}{4}\cos{(4A)} \\ &= \frac{3}{4} + \frac{1}{4}\cos{(4A)} \\ &= \frac{1}{4} \left[ 3 + \cos{(4A)} \right] \end{align*}
• Jun 16th 2013, 09:45 AM
Soroban
Re: Trigonometric function
Hello, Trefoil2727!

Quote:

$\displaystyle \text{2. Given that }A, B, C\text{ are the angles of a triangle,}$
$\displaystyle \text{show that: }\:\tan C\:=\: \frac{\tan A+\tan B}{\tan A\tan B-1}$

$\displaystyle \text{Since }\,A+B+C \:=\:180^o,$

. . $\displaystyle \text{we have: }\:C \:=\:180^o - (A + B)$

$\displaystyle \text{Then: }\;\;\tan C \;=\;\tan[180^o - (A+B)] \;=\; \frac{\tan180^o - \tan(A+B)}{1 - \tan180^o\tan(A+B)}$

. . . . . . . . . . $\displaystyle =\;\frac{0 - \tan(A+B)}{1 - 0\tan(A+B)} \;=\;-\tan(A+B)$

. . . . . . . . . . $\displaystyle =\;-\frac{\tan A + \tan B}{1-\tan A\tan B} \;=\;\frac{\tan A + \tan B}{\tan A\tan B - 1}$