# Thread: [SOLVED] finding the value of α

1. ## [SOLVED] finding the value of α

How do find the value of α in

sec(2α+6°)cos(5α+3°)=1

2. Originally Posted by somanyquestions
How do find the value of α in

sec(2α+6°)cos(5α+3°)=1
Use the sum formula for cosine

$\displaystyle \sec{(2a + 6^\circ)}\cos{(5a + 3^\circ)} = 1$

$\displaystyle \frac{\cos{(5a + 3^\circ)}}{\cos{(2a + 6^\circ)}} = 1$

$\displaystyle \frac{\cos{5a}\cos{3^\circ} - \sin{5a}\sin{3^\circ}}{\cos{2a}\cos{6^\circ} - \sin{2a}\sin{6^\circ}} = 1$.

Can you go from here?

3. hmm i don't know that rule..

4. $\displaystyle \cos{(\alpha \pm \beta)} = \cos{\alpha}\cos{\beta} \mp \sin{\alpha}\sin{\beta}$.

This is a standard identity.

5. I don't know "sec", can you tell me...

6. Actually I've thought of an easier way...

You have

$\displaystyle \frac{\cos{(5a + 3^\circ)}}{\cos{(2a + 6^\circ)}} = 1$

$\displaystyle \cos{(5a + 3^\circ)} = \cos{(2a + 6^\circ)}$

$\displaystyle 5a + 3^\circ = 2a + 6^\circ$

$\displaystyle 3a = 3^\circ$

$\displaystyle a = 1^\circ$.

7. it doesn't say anything about this in my book and we didn't go over this in class. strange.

8. $\displaystyle \sec{\theta} = \frac{1}{\cos{\theta}}$.

9. oh, thank you

10. Dear somanyquestions,

$\displaystyle sec(2a+6)^{o}cos(5a+3)^{o}=1$

$\displaystyle \frac{\cos{(5a + 3)}^o}{\cos{(2a + 6)^o}} = 1$

$\displaystyle cos(5a+3) = cos(2a+6)$

$\displaystyle \frac{\pi(5a+3)}{180}=2n{\pi}{\pm}\frac{{\pi}(2a+6 )}{180}$ ; $\displaystyle n\in{Z}$

By simplification you would get, $\displaystyle a = \frac{(360n-9)}{7}$ or $\displaystyle a = 120n+1$

Dear Prove It,

Your answer is a particular solution. For example $\displaystyle a = 121^o$ also satisfies the given trignometric equation.

Hope this helps.