# Proving and finding values.

• Nov 27th 2010, 06:31 PM
arccos
Proving and finding values.
I am having some problems with these questions,can anybody help?

2) It is given that sinA = 5/13 where 90<A<180(degrees) , and that cosB = -33/65 , where 180<B<270. Find the exact value of cos2(A+B)

How do i expand cos2(A+B) ? apparently when i use my calculator cos2(A+B) = cos(2A + 2B) but again,the expansion gets too complicated and i derived a six digit fraction which is a far cry from the answer (7/25)

3)A right pyramid has a square horizontal base ABCD of side 2x and a vertex O. Given that Angle AOB = $2\theta$ , show that the inclination of OA to the horizontal is $arcsin(\sqrt cos2\theta)$ and find the inclination of the plane OAB to the horizontal.

I can't think of any way to start on this problem where I am able to ultilize all of the information given.

1)(solved) If $A + B + C = \dfrac{\pi}{2} \,radians$ , show that $tanA tanB + tanB tanC + tanC tanA = 1$

For this I was thinking to make the pi/2 into 1 we have to introduce sine to A+B+C but it just gets more and more complicated when i try to expand everything and I feel that there should be a simpler way which I have overlooked.

Any help at all is appreciated
• Nov 27th 2010, 06:56 PM
harish21
1) $A+B+C=\dfrac{\pi}{2}$

$A = \dfrac{\pi}{2} - (B+C)$

$\tan (A) = \tan \bigg(\dfrac{\pi}{2} - (B+C)\bigg)$

Since $tan(\frac{\pi}{2} - x) = \cot (x)$

$\tan (A) = \cot (B+C)$

convert cot into tan and finish..
• Nov 27th 2010, 09:23 PM
Soroban
Hello, arccos!

Quote:

$\text{2) It is given that }\sin A = \frac{5}{13}\,\text{ where }90^o < A < 180^o$

$\text{and that }\cos B = -\frac{33}{65}\,\text{ where }180^o

$\text{Find the exact value of: }\:\cos2(A+B)$

You need these identities: . $\cos2\theta \;=\;2\cos^2\!\theta-1$

. . . . . . . . . . . . . . . . $\cos(A + B) \;=\;\cos A\cos B - \sin A\sin B$

We are given: . $\sin A \,=\,\frac{5}{13}$ in Quadrant 2.
. . Hence: . $\cos A \:=\:-\frac{12}{13}$

We are given: . $\cos B \,=\,-\frac{33}{65}$ in Quadrant 3.
. . Hence: . $\sin B \,=\,-\frac{56}{65}$

$\cos2(A+B) \;=\;2\cos^2(A+B) - 1$

. . . . . . . . . . $=\; 2\bigg[\cos A\cos B - \sin A\sin B\bigg]^2 - 1$

. . . . . . . . . . $=\; 2\bigg[\left(\text{-}\frac{12}{13}\right)\left(\text{-}\frac{33}{65}\right) - \left(\frac{5}{13}\right)\left(\text{-}\frac{56}{65}\right) \bigg] - 1$

. . . . . . . . . . $=\;2\bigg[\frac{396}{845} + \frac{280}{845}\bigg]^2 - 1 \;\;=\;\;2\bigg[\frac{676}{845}\bigg]^2 - 1$

. . . . . . . . . . $=\;\frac{913,952}{714,025} - 1 \;\;=\;\;\frac{199,927}{714,025} \;\;=\;\;\dfrac{7}{25}$

• Nov 27th 2010, 10:40 PM
arccos
thank you! i totally overlooked the -1