0
$\begingroup$

I have these two problems I'm working on!

First of the Double Angle Formula! This formula I attempted to do a lot but couldn't get to the identity! $$\sin^2 \theta \cdot \cos^2\theta = \tfrac18[1 - \cos(4 \theta)]$$

For Above question you can only use the following: \begin{align} \sin^2\theta &= \tfrac12 (1-\cos(2 \theta)) \\ \cos^2\theta &= \tfrac12(1 + \cos( 2 \theta )) \end{align}

And lastly this Sum And Difference Formula! I tried this one so much, I'm leaning toward it being impossible (it's obviously not.... because it's a question): $$\cos(a-b) \cdot \cos(a + b) = (\cos^2a - \sin^2b).$$

$\endgroup$
  • $\begingroup$ Formatting tips here. $\endgroup$ – Em. Apr 27 '16 at 21:51
  • $\begingroup$ Can you use other trigonometric identities? $\endgroup$ – wssbck Apr 27 '16 at 21:58
  • $\begingroup$ No @wssbck thats why this question was diffcult for me! For the first one you can only use Double Angle. And for the second one only the Sum/Diffrence Trignometric Identites :( . $\endgroup$ – amanuel2 Apr 27 '16 at 22:00
  • $\begingroup$ Your "double angle" formula is the same as the identity you want to prove. Could you correct this? $\endgroup$ – Alex R. Apr 27 '16 at 22:06
  • $\begingroup$ I Edited my question @AlexR. $\endgroup$ – amanuel2 Apr 27 '16 at 22:08
1
$\begingroup$

$$\sin \theta \cos \theta = \frac{1}{2}\sin 2\theta \tag{1}$$ $$\sin^2 2\theta = \frac{1}{2} (1- \cos 4\theta) \tag{2}$$ From (1) and (2), $$\sin^2 \theta \cos^2 \theta = \frac{1}{8}(1-\cos 4\theta)$$

$\endgroup$
1
$\begingroup$

For the second: \begin{align*} \cos(a-b)\cos(a+b) &= (\cos a\cos b + \sin a\sin b)(\cos a\cos b - \sin a \sin b) \\ &= \cos^2 a\cos^2 b - \sin^2 a\sin^2 b \\ &= \cos^2 a(1-\sin^2 b) - (1-\cos^2 a)\sin^2 b \\ &= \cos^2 a - \sin^2 b. \end{align*}

$\endgroup$
  • $\begingroup$ Nvm what i said earlier.. im just a dumass.. Thanks for anwsering! $\endgroup$ – amanuel2 Apr 27 '16 at 23:04
1
$\begingroup$

For the first identity,

$$\begin{align} {1\over8}(1-\cos4\theta)&={1\over8}(1-(2\cos^22\theta-1))\\ &={1\over4}(1-\cos^22\theta)\\ &={1\over4}\sin^22\theta\\ &={1\over4}(2\sin\theta\cos\theta)^2\\ &=\sin^2\theta\cos^2\theta \end{align}$$

For the second identity, see rogerl's answer. (I would do it the exact same way.)

$\endgroup$
0
$\begingroup$

$$\cos^2(\theta)\sin^2(\theta)=(1-\cos(2\theta)(1+\cos(2\theta))/4=(1-\cos^2(2\theta))/4.$$

$$1-\cos^2(2\theta)=1/2-\cos(4\theta)/2.$$

The identity should now follow.

$\endgroup$
  • $\begingroup$ What question did you do Alex? I am sooo confused now...? $\endgroup$ – amanuel2 Apr 27 '16 at 22:16
0
$\begingroup$

For the second one:

$$ \cos (a-b) \cdot \cos (a+b) = \\ = (\cos a \cdot \cos b + \sin a \cdot \sin b) \cdot (\cos a \cdot \cos b - \sin a \cdot \sin b) = \\ = \cos^2 a \cdot \cos^2 b - \sin^2 a \cdot \sin^2 b = \\ = \cos^2 a \cdot (1 - \sin^2 b) - \sin^2 b \cdot (1 - \cos^2 a) = \\ = \cos^2 a - \cos^2 a \cdot \sin^2 b - \sin^2 b + \sin^2 b \cdot \cos^2 a = \\ = \cos^2 - \sin^2 b $$

$\endgroup$
0
$\begingroup$

For the second formula, use the identity $$\cos x\cos y=\frac12(\cos(x-y)+\cos(x+y)),$$ which gives \begin{align*} \cos(a-b)\cos(a+b)&=\frac12(\cos 2b+\cos 2a)\\&=\frac12(1-2\sin^2b+2\cos^2a-1)\\&=\dotsm \end{align*}

$\endgroup$
0
$\begingroup$

For the first one: $$\\ \sin^2 \theta \cos^2 \theta = \frac{1-\cos 2\theta}{2}\cdot \frac{1+\cos 2\theta}{2}\\ \, =\frac{1-\frac{1+\cos 4\theta}{2}}{4}.$$

For the second one: $$\\ \quad \quad \cos(\alpha+\beta)=\cos \alpha \cos \beta - \sin \alpha \sin \beta \\\underline{\quad +\quad \cos(\alpha-\beta)=\cos \alpha \cos \beta + \sin \alpha \sin \beta \quad} \\ \quad \quad \quad \cos(\alpha+\beta)+\cos(\alpha-\beta)=2\cos \alpha \cos \beta.$$

Now put $\alpha=a-b,\ \beta=a+b$.

$\endgroup$
0
$\begingroup$

For the first, recall that: $$\sin x = \frac{ie^{-ix} - ie^{ix}}{2} \ ; \cos x = \frac{e^{-ix}+e^{ix}}{2}$$

Therefore, $\sin^2 \theta \cdot \cos^2 \theta$ equals:

$$\frac{1}{4} (ie^{-ix} - ie^{ix})^2 \cdot \frac{1}{4} (e^{-ix}+e^{ix})^2$$ $$\Rightarrow \frac{1}{4} \left(-e^{-2ix} -2 (-1) -e^{2ix} \right) \cdot\frac{1}{4} (e^{-2ix} + 2 + e^{2ix})$$ $$\Rightarrow \frac{1}{16} (-e^{-4ix}-2e^{-2ix}-1+2e^{-2ix}+4+2e^{2ix}-1-2e^{2ix}-e^{4ix})$$ $$ \Rightarrow \frac{1}{16} (-e^{-4ix} + 2 - e^{4ix}) $$ $$ \Rightarrow \frac{1}{8} (-\frac{1}{2}e^{-4ix} + 1 - \frac{1}{2}e^{4ix})$$ $$ \Rightarrow \frac{1}{8} \left(1 - \frac{1}{2}(e^{-4ix} + \frac{1}{2}e^{4ix}) \right)$$ $$\Rightarrow \frac{1}{8} \left( 1- \frac{1}{2} \cos(4x) \right)$$

$\endgroup$
0
$\begingroup$

We can also try to differentiate $\sin^2 \theta \cos^2 \theta - \frac{1}{8} \left(1 - \cos 4 \theta \right)$:

$$\frac{\mathrm d}{\mathrm d \theta} \left( \sin^2 \theta \cos^2 \theta - \frac{1}{8} \left(1 - \cos 4 \theta \right) \right)$$ $$= 2 \sin \theta \cos \theta \cdot \cos^2 \theta + \sin^2 \theta \cdot -2 \cos \theta \sin \theta - \frac{1}{2} \sin 4 \theta $$ $$= \sin 2\theta(\cos^2 \theta - \sin^2 \theta) - \sin 2 \theta (\cos 2 \theta) $$ $$= 0 $$

Now we need to make sure that the function is not a constant. Substituting $\theta = 0$ for example gives $0$, so $\sin^2 \theta \cos^2 \theta - \frac{1}{8} \left(1 - \cos 4 \theta \right) = 0 \Rightarrow \sin^2 \theta \cos^2 \theta =\frac{1}{8} \left(1 - \cos 4 \theta \right)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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