1. ## Trig Equations

Just need some help or pointers to solve these equations

Solve for 0 ≥ b ≥ 180

a) 3tanb - 5secb = 4

b) 2sinb - cosb = 2
i used auxillary angles for this question but i didnt get the right answer which was 90 and 143 degrees 8 minutes

Thankyou for any help =D

2. Originally Posted by THSKluv
i used auxillary angles for this question but i didnt get the right answer which was 90 and 143 degrees 8 minutes
If you post your working so far we can see where abouts you went wrong.

3. 2sinb - cos b = R sin ( b+a)
= R (sin b cos a + sin a cos b)
2 = R cos a
-1 = R sin a

R = root 5

cos a = 2/root5
sin a = (-1)/root 5

therefore a = 333 degrees 26 minutes
root5 sin ( b + 333 degrees 36 minutes) = 2
b + 333degrees 26 minutes = 63 degreees 26 minutes or 116 degrees 34 minutes

yeaa thats my working out, and that DEF. doesnt equal the answer given.
><"" T.T !!!

4. Originally Posted by THSKluv
2sinb - cos b = R sin ( b+a)
= R (sin b cos a + sin a cos b)
yeaa thats my working out, and that DEF. doesnt equal the answer given.
><"" T.T !!!
I'm not sure how you have got the above equation?

Here's how I would do it.

$\displaystyle 2\sin{b} - \cos{b} = 2$

$\displaystyle 2\sin{b} - 2 = \cos{b}$, if we now square both sides

$\displaystyle (2\sin{b} - 2)^2 = \cos^2{b}$, now using the fact that $\displaystyle \sin^2{b} + \cos^2{b} = 1$, we can write this as

$\displaystyle (2\sin{b} - 2)^2 = 1 - \sin^2{b}$

$\displaystyle 4\sin^2{b} - 8\sin{b} + 4 = 1 - \sin^2{b}$

$\displaystyle 5\sin^2{b} - 8\sin{b} + 3 = 0$.

This can now be factorised using the quadratic formula.

5. Originally Posted by THSKluv
Just need some help or pointers to solve these equations

Solve for 0 ≥ b ≥ 180

a) 3tanb - 5secb = 4

b) 2sinb - cosb = 2
i used auxillary angles for this question but i didnt get the right answer which was 90 and 143 degrees 8 minutes

Thankyou for any help =D
a) $\displaystyle 3\tan b-5\sec b=4$

$\displaystyle \Rightarrow \frac{3\sin b}{\cos b}-\frac5{\cos b}=4$

$\displaystyle \Rightarrow \frac{3\sin b-5}{\cos b}=4$

$\displaystyle \Rightarrow 3\sin b-5=4\cos b$

Squaring,

$\displaystyle 9\sin^2b+25-30\sin b=16\cos^2 b$

$\displaystyle \Rightarrow 9\sin^2b+25-30\sin b=16(1-\sin^2b)$

$\displaystyle \Rightarrow 9\sin^2b+25-30\sin b=16-16\sin^2b$

$\displaystyle \Rightarrow 25\sin^2b-30\sin b+9=0$

$\displaystyle \Rightarrow \sin b=\frac{30\pm\sqrt{900-900}}{50}$

$\displaystyle \Rightarrow \sin b=\frac{30}{50}$

$\displaystyle \Rightarrow b=\sin^{-1}\left(\frac{3}{5}\right)$

6. ## Trig equations

Hello everyone

The problem with the solutions offered so far is that squaring both sides of an equation nearly always runs the risk of introducing unwanted answers, which then need to be eliminated.

For example, Craig's equation

$\displaystyle 5\sin^2{b} - 8\sin{b} + 3 = 0$

can be factorised to give

$\displaystyle (5\sin b -3)(\sin b -1) = 0$

$\displaystyle \Rightarrow \sin b = \tfrac35,\, 1$

$\displaystyle \Rightarrow b = 36.87^o$ (among other possible answers), which just doesn't satisfy the original equation.

$\displaystyle \sin b = \tfrac35$ suffers from the same problem: $\displaystyle 3\tan36.87^o - 5 \sec36.87^o = -4$, not $\displaystyle +4$.

THSKluv's original approach is the right one to pursue, but it's easier if you use
$\displaystyle \sin(b-a)$, not $\displaystyle \sin(b+a)$:

$\displaystyle 2\sin b -\cos b = R\sin(b-a) = R(\sin b \cos a - \cos b \sin a)$

$\displaystyle \Rightarrow R = \sqrt5$ and $\displaystyle a = \tan^{-1}(\tfrac12) = 26.57^o$

$\displaystyle \Rightarrow \sqrt5\sin(b - a) = 2$

$\displaystyle \Rightarrow b - a = \sin^{-1}\Big(\frac{2}{\sqrt5}\Big) = 63.43, 116.57, ...$

$\displaystyle \Rightarrow b = 90^o, 143.1^o, ...$

The other equation can be solved in the same way. Using great_math's original approach, you get:

$\displaystyle 3\sin b - 4\cos b = 5$

Now continue, by writing
$\displaystyle 3\sin b - 4\cos b=R\sin(b-a) = ...$

Can you complete it now?