# Show that $\sin^2 \theta \cdot \cos^2\theta = (1/8)[1 - \cos(4 \theta)]$.

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

• Formatting tips here. – Em. Apr 27 '16 at 21:51
• Can you use other trigonometric identities? – wssbck Apr 27 '16 at 21:58
• 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 :( . – amanuel2 Apr 27 '16 at 22:00
• Your "double angle" formula is the same as the identity you want to prove. Could you correct this? – Alex R. Apr 27 '16 at 22:06
• I Edited my question @AlexR. – amanuel2 Apr 27 '16 at 22:08

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

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*}

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

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.)

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

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

The identity should now follow.

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

For the second one:

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

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*}

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

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:

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

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 $$\sin^2 \theta \cos^2 \theta - \frac{1}{8} \left(1 - \cos 4 \theta \right) = 0 \Rightarrow \sin^2 \theta \cos^2 \theta =\frac{1}{8} \left(1 - \cos 4 \theta \right)$$.