# Trig help

• Jul 14th 2014, 11:51 PM
rainbowstar
Trig help
Hey guys,
I am confused with these two questions (8 and 9). I have no idea where to begin off.
These questions are where one trigonometric function is given and find another.

It would really be help if you could post how to work out this questions.

Thanks
Attachment 31275
• Jul 15th 2014, 01:20 AM
romsek
Re: Trig help
you can note that

$\sin(\theta+90^\circ)=\sin(\theta)\cos(90^\circ)+ \cos(\theta)\sin(90^\circ)=\cos(\theta)$

and that

$\cos(\theta+90^\circ)= \cos(\theta)\cos(90^\circ)-\sin(\theta)\sin(90^\circ)=-\sin(\theta)$

so

$\tan(\theta + 90^\circ)=\dfrac{\cos(\theta)}{-\sin(\theta)}$

$\theta$ is obtuse so

$\sin(\theta) > 0$ and $\cos(\theta)<0$

so

$\cos(\theta)=-\sqrt{1-sin^2(\theta)}=-\sqrt{1-k^2}$

and thus

$\tan(\theta+90^\circ)=\dfrac{-\sqrt{1-k^2}}{-k}=\dfrac{\sqrt{1-k^2}}{k}$

for (9)

$\sec^2(\theta)=1+\tan^2(\theta)$

$\sec^2(\theta)-1=a^2-\dfrac 1 2 + \dfrac 1 {16a^2}=\left(a-\dfrac 1 {4a}\right)^2$

so

$\tan(\theta)=a - \dfrac 1 {4a}$

$\sec(\theta)+\tan(\theta)=\left(a + \dfrac 1 {4a}\right) + \left(a - \dfrac 1 {4a}\right)=2a$
• Jul 15th 2014, 01:45 AM
rainbowstar
Re: Trig help
Quote:

Originally Posted by romsek
you can note that

$\sin(\theta+90^\circ)=\sin(\theta)\cos(90^\circ)+ \cos(\theta)\sin(90^\circ)=\cos(\theta)$

and that

$\cos(\theta+90^\circ)= \cos(\theta)\cos(90^\circ)-\sin(\theta)\sin(90^\circ)=-\sin(\theta)$

so

$\tan(\theta + 90^\circ)=\dfrac{\cos(\theta)}{-\sin(\theta)}$

$\theta$ is obtuse so

$\sin(\theta) > 0$ and $\cos(\theta)<0$

so

$\cos(\theta)=-\sqrt{1-sin^2(\theta)}=-\sqrt{1-k^2}$

and thus

$\tan(\theta+90^\circ)=\dfrac{-\sqrt{1-k^2}}{-k}=\dfrac{\sqrt{1-k^2}}{k}$

for (9)

$\sec^2(\theta)=1+\tan^2(\theta)$

$\sec^2(\theta)-1=a^2-\dfrac 1 2 + \dfrac 1 {16a^2}=\left(a-\dfrac 1 {4a}\right)^2$

so

$\tan(\theta)=a - \dfrac 1 {4a}$

$\sec(\theta)+\tan(\theta)=\left(a + \dfrac 1 {4a}\right) + \left(a - \dfrac 1 {4a}\right)=2a$

Thankyou very much. That helped a lot!