Prove the following trignometric identities
√(1-cos^2 θ)/(cos^2 θ) = tanθ
(3π/2 + ϕ) = sin ϕ
For the first one, one of the
Pythagorean Identities says that \(\displaystyle sin^2(\theta) + cos^2(\theta)=1\). What does that tell you about the value of \(\displaystyle 1-cos^2(\theta)\)? Next, because of the periodicity of sine and cosine, we can safely assume that \(\displaystyle 0 \le \theta \le 2\pi\) (or \(\displaystyle 0^\circ \le \theta \le 360^\circ\)), meaning that theta is always positive. Returning to the rules you learned in algebra, what can you say about \(\displaystyle \sqrt{f^2(x)}\) if x is guaranteed to be positive? Finally, recall the basic identity that \(\displaystyle tan(\theta)=\dfrac{sin(\theta)}{cos(\theta)}\).
As for the second "identity." that's not really an identity. An identity means that the equation must hold for all values. If we plug in, say, pi/2 (90 degrees), we can see the equation doesn't hold:
\(\displaystyle \dfrac{3\pi}{2} + \dfrac{\pi}{2} = sin \left(\dfrac{\pi}{2} \right) \implies 2\pi = 1\)
Oops. That's not true. 6.28... does not equal 1! Perhaps revisit the problem and clarify any typos you (or possibly the book) may have made.
Solve the following trigonometric equations in the range 0 deg to 360 deg
2sec^2(ϕ) + 5 tan(ϕ) = 3
Here you'll want to use another one of the Pythagorean Identities. Namely \(\displaystyle sec^2(\phi) = tan^2(\phi)+1\). Making that substitution leaves you with a quadratic. Maybe make another substitution and let \(\displaystyle u=tan(\phi)\). Can you solve that quadratic in terms of u? Remember to back-substitute \(\displaystyle tan(\phi)\) when you're done.
Prove the hyperbolic identities
coth(x) = 2cosech(2x) + tan(x)
This is another case where what you've posted isn't actually an identity. Again, you'll need to review and correct any errors.
Given 5e^x - 4e^-x =A sinh x + B cosh x, Find A and B
For this problem, you'll need to know the definitions of hyperbolic sine and hyperbolic sine. Both those functions are defined in terms of e, as:
\(\displaystyle sinh(x)=\dfrac{e^x-e^{-x}}{2}\) and \(\displaystyle cosh(x)=\dfrac{e^x+e^{-x}}{2}\)
From the above, it follows that:
\(\displaystyle sinh(x)+cosh(x)=e^x\)
How do you think you might use this information to help solve the problem? As a hint, I'd start with the left-hand side of the equation and leave the right-hand side alone for now.