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2 years ago

Help please!!! I don't understand

Mathematics

1 answer:

zaharov [31]2 years ago

6 0

$\left|x-\dfrac 12\right| = 3\dfrac 23\\\\\\\implies \left|x- \dfrac 12 \right| = \dfrac{11}3\\\\\\\implies x-\dfrac 12 = \dfrac{11}3 ~~~~~~~~~ \text{or}~~~~~~~ x - \dfrac 12 = - \dfrac {11}3\\\\\\\implies 2x -1 = \dfrac{22}3 ~~~~~~~~ \text{or}~~~~~~~~2x-1 =-\dfrac{22}3\\$

$\implies x =\dfrac 12\left(\dfrac{22}3 +1 \right) ~~ \text{or}~~~~~~~~ x = \dfrac 12 \left(1-\dfrac{22}3\right)}\\\\\\\implies x = \dfrac{25}6~~~~~~~~~~~~~~~ \text{or}~~~~~~~~ x=-\dfrac{19}6\\\\\\\implies x = 4\dfrac 16 ~~~~~~~~~~~~~~~ \text{or}~~~~~~~~x = -3\dfrac16$

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$\theta$ is in quadrant I, so $\cos\theta>0$ .

$x$ is in quadrant II, so $\sin x>0$ .

Recall that for any angle $\alpha$ ,

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Then with the conditions determined above, we get

$\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35$

and

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Now recall the compound angle formulas:

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$

$\sin2\alpha=2\sin\alpha\cos\alpha$

$\cos2\alpha=\cos^2\alpha-\sin^2\alpha$

as well as the definition of tangent:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$

Then

1. $\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}$

2. $\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}$

3. $\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}$

4. $\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}$

5. $\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}$

6. $\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7$

7. A bit more work required here. Recall the half-angle identities:

$\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2$

$\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2$

$\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}$

Because $x$ is in quadrant II, we know that $\dfrac x2$ is in quadrant I. Specifically, we know $\dfrac\pi2$ , so $\dfrac\pi4$ . In this quadrant, we have $\tan\dfrac x2>0$ , so

$\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac32$

8. $\sin3\theta=\sin(\theta+2\theta)=\dfrac{44}{125}$

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