Exact value of Cos(Alpha-Beta)

**Neversh** · June 15th 2009, 02:22 PM

cosalpha = sqrt(3)/4 in quadrant I
cosbeta = -(sqrt2)/3 in quadrant II

Find the exact value of cos(alpha - beta)

**pickslides** · June 15th 2009, 04:26 PM

$cos(a-b) = cos(a)cos(b)+sin(a)sin(b)$

To find the values for sin(a) and sin(b) use $sin^2(x)+cos^2(x)=1$

**Soroban** · June 15th 2009, 04:46 PM

Hello, Neversh!

Given: . $\begin{array}{c}\cos\alpha \:=\: \dfrac{\sqrt{3}}{4}\;\text{ in Quadrant 1} \\ \\[-3mm]<br /> \cos\beta \:=\: \text{-}\dfrac{\sqrt{2}}{3}\;\text{ in Quadrant 2} \end{array}$

Find the exact value of: . $\cos(\alpha - \beta)$

We know: . $\sin^2\!\theta + \cos^2\!\theta \:=\:1 \quad\Rightarrow\quad \sin\theta \:=\:\pm\sqrt{1-\cos^2\!\theta}$

Since ${\color{blue}\cos\alpha = \frac{\sqrt{3}}{4}}$ , then $\sin\alpha \:=\:\pm\sqrt{1 - \left(\tfrac{\sqrt{3}}{4}\right)^2} \:=\:\pm\sqrt{1 - \tfrac{3}{16}} \:=\:\pm\sqrt{\tfrac{13}{16}} \:=\:\pm\tfrac{\sqrt{13}}{4}$

. . $\alpha$ is in Quadrant 1, so: . ${\color{blue}\sin\alpha \:=\:\frac{\sqrt{13}}{4}}$

Since ${\color{blue}\cos\beta = \text{-}\frac{\sqrt{2}}{3}}$ , then $\sin\beta\:=\:\pm\sqrt{1 - \left(\text{-}\tfrac{\sqrt{2}}{3}\right)^2} \:=\:\pm\sqrt{1-\tfrac{2}{9}} \:=\:\pm\sqrt{\tfrac{7}{9}} \:=\:\pm\tfrac{\sqrt{7}}{3}$

. . $\beta$ is in Quadrant 2, so: . ${\color{blue}\sin\beta \:=\:\frac{\sqrt{7}}{3}}$

We know: . $\cos(\alpha - \beta ) \;=\;\cos\alpha\cos\beta + \sin\alpha\sin\beta$

Substitute the above four values:

. . $\cos(\alpha - \beta) \;=\;\left(\tfrac{\sqrt{3}}{4}\right)\left(\text{-}\tfrac{\sqrt{2}}{3}\right) + \left(\tfrac{\sqrt{13}}{4}\right)\left(\tfrac{\sqr t{7}}{3}\right) \;=\;\text{-}\tfrac{\sqrt{6}}{12} + \tfrac{\sqrt{91}}{12}$

Therefore: . $\cos(\alpha - \beta) \;=\;\frac{\sqrt{91} - \sqrt{6}}{12}$

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