trigonometry, trigonometric identities, hyperbolic functions

[jimkeller1993]

i have been off college for a couple of weeks and have missed relevant information on an assignment, i have never done this before and i am completely clueless at to where to go, can anybody show me how to tackle any of these

A)

Prove the following trignometric identities

√(1-cos^2 θ)/(cos^2 θ) = tanθ

(3π/2 + ϕ) = sin ϕ


B)

solve the following trigonometric equations in the range 0 deg to 360 deg

2 sec ^ 2 ϕ + 5 tan ϕ = 3

C)

Prove the hyperbolic identities

coth x ≡ 2 cosech 2 x + tan x

D)

given 5e^x - 4e^-x ≡A sinh x + B cosh x

Find A and B

 

[ksdhart2]

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.

 

[jimkeller1993]

thank you very much for taking the time to review my post
In terms of attempting the questions:

i have completed the first one part a)
come up short on part b) the second one cos (3pi/2 + t) = sin t Cannot find the correct stages/method

i have the correct quadratic formula for the second question and have two answers for that

for the third one i have got as far as cothx= 1+sin^2x / sinhxcoshx but am struggling as to the next stage of this equation

have completed the 4th question with a value for A and B

Any guidance on these questions is greatly appreciated

 

[ksdhart2]

Okay, so if I'm understanding you correctly, you've solved all but two of the problems. I see you gave a correction to one of them, so it's now:

$\displaystyle \displaystyle cos \left( \frac{3\pi}{2} + \phi \right) = sin(\phi)$

Now, this is an identity. To solve this, try applying the known identity $\displaystyle \displaystyle cos \left(\phi - \dfrac{\pi}{2} \right)=sin(\phi)$, along with the fact that the period of both sine and cosine is 2pi.

Now the other unsolved problem is $\displaystyle coth(x) = 2cosech(2x) + tan(x)$, but, as I mentioned in my last post, this isn't an identity. It can be solved for specific values of x, if that's the goal, but you stated that the instructions were to "Prove the hyperbolic identity." As it stands, it's literally unsolvable due to not being an identity. You'll need to find out what the actual problem statement for this one is and correct it before you'll be able to proceed.