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Let $p=\sin 24^\circ$

  1. Then what would $\cos (24^\circ)$ be in terms of $p$?

  2. What would $\sin (168^\circ) \cdot \sin(-78^\circ)$ be in terms of $p$?

I'm not sure how to approach these as we have only started this section.


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In the future, if you want to ask two questions, please ask them in separate posts. – MJD May 13 '13 at 14:44
If this is homework it should really have the homework tag. – Clive Newstead May 13 '13 at 15:24

4 Answers 4

You know that $\sin^2+\cos^2=1$. As $\sin(24)=p$ you have $p^2 + \cos^2(24)=1$. Solve this equation for $\cos(24)$. For the second part you can use some trigonometric theorems. For example $$ \sin(\alpha) \cdot \sin(\beta) = -\frac{1}{2} (\cos(\alpha+\beta) -\cos(\alpha - \beta))$$

This gives you $$\sin(168^\circ) \cdot \sin(-78^\circ)=-\frac{1}{2} ( \cos(90^\circ)- \cos(246^\circ))$$ We know that $\cos(90^\circ)=0$ and that $\cos(x+180^\circ)=-\cos(x)$ hence we have $$-\frac{1}{2} \cos(66^\circ)=-\frac{1}{2}\cos(90^\circ-24^\circ)=-\frac{1}{2} \sin(24^\circ)$$

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I should mention that the sign of $\sin(168^\circ)\times\sin(-78^\circ)$ is wrong. – robjohn May 13 '13 at 15:31
Oh how I hate those signs. – Dominic Michaelis May 13 '13 at 15:34
@robjohn thanks and fixed – Dominic Michaelis May 13 '13 at 15:34

Part 1

Use the identity \begin{equation*} \sin^2A+\cos^2A=1\\ \cos^2 A={1-\sin^2 A} \end{equation*} so \begin{align*} \cos^2 24^\circ&={1-\sin^2( 24^\circ) }\\ \\ \\cos^2 24^\circ&={1-p^2 } \\ \\ \cos24^\circ&=\sqrt{1-p^2 } \end{align*} Here we only take positive value of square root because $24^\circ$ is in I quaderent and cos ratio in quaderent I is positive.

Part 2

\begin{align*} \sin 168^\circ \cdot \sin(-78^\circ) &=-\frac{2\sin 168^\circ\cdot\sin78^\circ}{2} \end{align*} Using the formula ${2\sin A\cdot\sin B}=\cos (A-B)-\cos(A+B)$ \begin{align*} \sin(168^\circ)\cdot \sin(-78^\circ)&=-\frac{\cos 90^\circ-\cos 246^\circ}{2} \\ &=\frac{\cos 246^\circ}{2}\\ &=\frac{\cos(270^\circ-24^\circ)}{2}\\ &=-\frac{\sin 24^\circ}{2}\\ &=-\frac{p}{2} \end{align*}

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if somebody downvote then pls mention my errors.Thanks – iostream007 May 13 '13 at 15:29
I didn't downvote, but you got the sign of $\sin(168^\circ)\times\sin(-78^\circ)$ wrong. – robjohn May 13 '13 at 15:30
I did not downvote, but your formatting could is not that nice, do you need help improving it ? – Dominic Michaelis May 13 '13 at 15:31
@iostream May I fix the formatting ? – Dominic Michaelis May 13 '13 at 15:35
@DominicMichaelis NO it is perfect.Thanks for it – iostream007 May 13 '13 at 16:35

enter image description here

My favorite method

$$\sin \theta = \cfrac{\text {opposite}}{\text{hypothenus}}=\cfrac p1,\quad \quad \cos\theta = \cfrac{\text {adjacent}}{\text{hypothenus}}$$ What is "adjacent"? By Pthagoras', we have $\quad \text{adjacent}^2+\text{opposite}^2=\text{hypothenus}^2$ so $$ \text{adjacent}^2 + p^2=1 \implies \text{adjacent}=\sqrt{1-p^2}$$ et voila! $$\cos\theta = \cfrac{\text {adjacent}}{\text{hypothenus}}=\cfrac{\sqrt{1-p^2}}{1}=\sqrt{1-p^2}$$

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Why did you write hypothenus and opposite in italics on the graphic and he variables in recto ? – Dominic Michaelis May 13 '13 at 15:54

Part 2

Method 1:

$\sin168^\circ=\sin(180^\circ-168^\circ)$ ( as $\sin(180^\circ-\theta)=\sin\theta$)

So, $\sin168^\circ=\sin12^\circ$

and $\sin(-78^\circ)=-\sin78^\circ$ as $\sin(-\theta)=-\sin\theta$

So, $\sin(-78^\circ)=-\sin(90^\circ-12^\circ)=-\cos12^\circ$

$$\implies \sin168^\circ\cdot\sin(-78^\circ)=\sin12^\circ\cdot(-\cos12^\circ)=-\frac{\sin (2\cdot12^\circ)}2=-\frac{\sin24^\circ}2$$

Method 2:

$\sin168^\circ=\sin(90^\circ+78^\circ)=\cos78^\circ$ as $\sin(90^\circ+\theta)=\cos\theta$

Again, $\cos78^\circ=\cos(-78^\circ)$ as $\cos(-\theta)=\cos\theta$

$$\implies \sin168^\circ\cdot\sin(-78^\circ)=\cos(-78^\circ)\cdot\sin(-78^\circ)=\frac{\sin 2(-78^\circ)}2$$ $$=\frac{\sin(24^\circ-180^\circ)}2=-\frac{\sin24^\circ}2$$ as $\sin(\theta-180^\circ)=-\sin(180^\circ-\theta)=-\sin\theta$

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