Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
  1. I need to show that $\sin 4a = 0$ if $\cos a = 0$.

    I am not sure how to do this really.

    I know I can take $\sin^2 x + \cos^2 x = 1$ but I don't think that helps.

  2. I was also suppose to find the period of $\sin 2x + \cos 5x$, I know the answer because I graphed it but is there a really easy formula or way to look at it and see the answer? I came up with $2\pi$ because that is where $10\pi/5$ and $2\pi$ will meet. ($\pi/5$ and $\pi$)

EDIT: I think I just got it actually, if $\cos a$ is 0 that means $a$ is a multiple of 90 degrees so 4 times that is 360, whose is sin 0.

share|improve this question
2  
Please ask your second problem as an entirely new question instead of as an edit to this one. –  Ben Alpert Jun 22 '11 at 1:54
add comment

5 Answers 5

up vote 5 down vote accepted

What other identities have you tried? Perhaps the double angle identity will be helpful here: $ \sin(2a) = 2\sin(a)\cos(a) $ (This is of course for the first question, as Ben Alpert said, you should consider posting the second part as a separate question.)

share|improve this answer
    
That makes sense actually, sin2a = sincos is that accurate? Not that it matters, but since sin is 4a I just divide by 2 and get that. –  Adam Jun 22 '11 at 2:18
1  
you cannot simply divide by two in this case, because the 4a is being passed into the function. However, you can reapply the double angle identity, $ sin(4a) = sin(2(2a)) $. –  Alpha Jun 22 '11 at 2:21
2  
@Adam: you need to be more careful-this shows up repeatedly. $\sin2a=2\sin a \cos a$. You lost the factor $2$ and not putting the $a$ on the right is a good way to overlook the fact that you have to apply the double angle identities again-both for $\sin$ and $\cos$ –  Ross Millikan Jun 22 '11 at 2:52
add comment

You got it in your edit. Just be careful that $\cos x=0$ means that $x$ is an odd multiple of $90^\circ$ (the cosine of just any integer multiple of $90^\circ$ need not be $0$) and so $4x$ is a multiple of $360^\circ$ and so $\sin x=0$.

share|improve this answer
add comment

Starting with the addition identities:

$$\sin (a+b)=\sin a\cdot \cos b+\cos a\cdot \sin b,$$

$$\cos (a+b)=\cos a\cdot \cos b-\sin a\cdot \sin b,$$

we get (setting $a=b$)

$$\sin (2a)=2\sin a\cdot \cos a,$$

$$\cos (2a)=\cos ^{2}a-\sin ^{2}a.$$

Hence

$$\sin (4a)=2\sin (2a)\cdot \cos (2a)=2\cdot 2\sin a\cdot \cos a\left( \cos ^{2}a-\sin ^{2}a\right). $$

If $\cos a=0$, then

$$2\cdot 2\sin a\cdot \cos a\left( \cos ^{2}a-\sin^{2}a\right) =0,$$

and so is $\sin (4a)=0$.

This is not the minimal calculation, but it is very automatic.

share|improve this answer
add comment

For your second question...

In general, the graph of $y=f(2x)$ looks like the graph of $y=f(x)$, except that it is compressed horizontally around the $y$-axis by a factor of $2$ (the function "goes through" the values of $x$ twice as fast; so $f$ takes all values it used to take between, say, $0$ and $2$, but does it between $0$ and $1$; so what used to "take" all the way to $x=2$ to graph now needs to be graphed between $0$ and $1$, hence "twice as fast").

Likewise, the graph of $y=f(3x)$ is like the graph of $y=f(x)$, compressed by a factor of $3$; $y=f(4x)$ is like the graph of $y=f(x)$ compressed by a factor of $4$, etc. a

That means that the graph of $y=sin(2x)$ will look just like the graph of $y=\sin(x)$, but compressed by a factor of $2$. Since $y=\sin(x)$ takes $2\pi$ to do a full cycle, then $y=\sin(2x)$ will take half as long (will do the cycle "twice as fast"). So the period is $\pi$ instead of $2\pi$.

Similarly, the graph of $y=\cos(5x)$ is like the graph of $y=\cos(x)$, but compresssed by a factor of $5$; it finishes a cycle five times as fast as the regular cosine does; so it does a cycle every $\frac{2\pi}{5}$.

If you have one function repeating every $\pi$, and one function repeating every $\frac{2\pi}{5}$, then they will "sync back up" at $2\pi$, because that is the smallest integer multiple of both $\pi$ and $\frac{2\pi}{5}$, as you surmise.

Alternatively: if $y=g(x)$ is periodic with period $P$, that means that $g(x+P)=g(x)$ for all $x$. What is the period of $y=g(2x)$? Well, if you instead of $x$ you plug in $x+\frac{P}{2}$, you get: $$y = g\left( 2\left(x+\frac{P}{2}\right)\right) = g(2x + P) = g(2x)$$ so $\frac{P}{2}$ is a period for $y=g(2x)$. Generally, if $y=g(x)$ has period $P$, then $y=g(kx)$ has period $\frac{P}{k}$ (can you see why?).

share|improve this answer
add comment

First question - haven't the rep for a comment. Apply the double angle formula twice for sin, and you get an expression sin(4a) = 4cos(a)sin(a)cos(2a) the first factor of which is zero - you don't need to expand the cos(2a) bit.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.