proving sin and cos identities

**allywallyrus** · March 23rd 2009, 07:24 AM

i'm not sure im doing this right and im having trouble understanding this lesson
what i have is:
$\cos\theta+\sin\theta \tan\theta = \displaystyle{\frac{1}{\cos\theta}}$
$\cos\theta+\sin\theta\displaystyle{\frac{\sin\thet a}{\cos\theta}}$
\cos\theta+\displaystyle{\frac{\sin^2\theta\}{\cos \theta}}
$\displaystyle{\frac{\cos^2\theta+\sin^2\theta}{\co s\theta}}$
which would give me the correct answer but im worried i've done something incorrectly with the fractions
also i'm kind of new to this latex thing, but the third line was giving me a syntax error and i don't know what i did wrong exactly so i took the tags off

**e^(i*pi)** · March 23rd 2009, 07:28 AM

Originally Posted by allywallyrus

i'm not sure im doing this right and im having trouble understanding this lesson
what i have is:
$\cos\theta+\sin\theta \tan\theta = \displaystyle{\frac{1}{\cos\theta}}$

$\cos\theta+\sin\theta\displaystyle{\frac{\sin\thet a}{\cos\theta}}$

$cos(\theta)+ \frac{sin^2(\theta)}{cos(\theta)}$

$\displaystyle{\frac{\cos^2\theta+\sin^2\theta}{\co s\theta}}$

which would give me the correct answer but im worried i've done something incorrectly with the fractions
also i'm kind of new to this latex thing, but the third line was giving me a syntax error and i don't know what i did wrong exactly so i took the tags off

Nope your maths is good, you just need to use the Pythagorean identity ( $sin^2(x) + cos^2(x) = 1$ ) to get to the final answer

**allywallyrus** · March 23rd 2009, 07:38 AM

oh good, thats what i was hoping to hear

and i dont suppose there is any rule against bumping this thread with new questions a little later on? i will probably have a couple more

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