Trigonometry Formulae - Part 4

Geometry Level 2

If $\tan (\alpha )=2$ , then $\dfrac{\sin (\alpha )+\cos (\alpha )}{3 \sin (\alpha )-2 \cos (\alpha )} = \dfrac ab$ , where $a$ and $b$ are coprime positive integers. Find $a+b$ .

The answer is 7.

2 solutions

Chew-Seong Cheong
May 20, 2019

$\begin{aligned} Q & = \frac {\sin \alpha + \cos \alpha}{3\sin \alpha - 2\cos \alpha} & \small \color{#3D99F6} \text{Divide up and down by }\cos \alpha \\ & = \frac {{\color{#3D99F6}\tan \alpha}+1} {3{\color{#3D99F6}\tan \alpha}-2} & \small \color{#3D99F6} \text{Since }\tan \alpha = 2 \\ & = \frac {{\color{#3D99F6}2}+1} {3{\color{#3D99F6}(2)}-2} \\ & = \frac 34 \end{aligned}$

Therefore, $a+b = 3+4 = \boxed 7$ .

@Alice Smith , you don't need to use \displaystyle for fractions, $\tan x$ , $\sin x$ and $\cos x$ . For fractions, just use \dfrac \pi 2 $\dfrac \pi 2$ . Similarly for \binom nk $\binom nk$ , \dbinom nk $\dbinom nk$ . \displaystyle is used for \int _0^\frac \pi 2, \sum _{k=0}^\infty and \ lim _{x \to infty} $\displaystyle \int_0^\frac \pi 2, \sum_{k=0}^\infty, \lim_{x \to \infty}$ , Without \displaystyle $\int _0^\frac \pi 2, \sum_{k=0}^\infty, \lim_{x \to \infty}$

Chew-Seong Cheong - 2 years ago

Thanks for mentioning that:)

Alice Smith - 2 years ago

Chris Lewis
May 20, 2019

Divide the numerator and denominator by $\cos \alpha$ (which, since $\tan \alpha =2$ , is not zero) to leave

$\frac{\tan \alpha + 1}{3\tan \alpha - 2} = \frac{3}{4}$

so $a+b=\boxed7$ .

Trigonometry Formulae - Part 4

The answer is 7.

2 solutions

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