Trigonometry Question (1 Viewer)

Ambility · Aug 27, 2015

$$Express $\cos{x}+7\sin{x}=5$ in the form $A\cos{(x+\alpha)}$. Hence, solve $\cos{x}+7\sin{x}=5$, 0^{\circ}\leq x\leq360^{\circ}$

I can express it in as a sine, but how do I get it to a cosine?

InteGrand · Aug 27, 2015

Ambility said:
$$Express $\cos{x}+7\sin{x}=5$ in the form $A\cos{(x+\alpha)}$. Hence, solve $\cos{x}+7\sin{x}=5$, 0^{\circ}\leq x\leq360^{\circ}$

I can express it in as a sine, but how do I get it to a cosine?

$\cos x + 7\sin x \equiv A\cos (x+\alpha)$

$$\Rightarrow \cos x + 7\sin x \equiv A\cos x \cos \alpha - A\sin x \sin \alpha$ (compound angle formula).$

$Equating coefficients of $\cos x$ and $\sin x$, we have$

$$A\cos \alpha =1$ \ \ (1)$

$and$

$$A\sin \alpha =-7$. \ \ (2)$

$Squaring and adding, $A^2 (\sin^2 \alpha + \cos^2 \alpha) = 49+1 = 50\Rightarrow A=\sqrt{50}$.$

$From $(1)$ and $(2)$, we have that $\alpha$ is in the fourth quadrant, so from $(1)$, $\cos \alpha = \frac{1}{\sqrt{50}}\Rightarrow \alpha = -\cos^{-1} \left( \frac{1}{\sqrt{50}}\right)$.$

$Therefore, $\cos x + 7\sin x \equiv \sqrt{50}\cos \left(x-\cos^{-1} \left( \frac{1}{\sqrt{50}}\right) \right)$.$

InteGrand · Aug 27, 2015

Ambility · Aug 27, 2015

Awesome, thanks. Why do you use the triple-lined equal sign?

InteGrand · Aug 27, 2015

Ambility said:
Awesome, thanks. Why do you use the triple-lined equal sign?

That means "identically equal to", meaning that we want that equation to be true for all values of x. It's not a big deal though if you use a normal equals sign (=) in the exam.

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Trigonometry Question (1 Viewer)

Ambility

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InteGrand

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InteGrand

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Ambility

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InteGrand

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