Question

Let $\alpha$ and $\beta$ be first quadrant angles with $\cos(\alpha) = \frac{\sqrt{3}}{2}$ and $\sin(\beta) = \frac{\sqrt{6}}{6}$. Find $\cos(\alpha + \beta)$. $\cos(\alpha + \beta) = $ No solution $\frac{\sqrt{3}}{2}\frac{\sqrt{30}}{6} - \frac{\sqrt{1}}{2}\frac{\sqrt{6}}{6} \approx 0.58644526981016$

          Let $\alpha$ and $\beta$ be first quadrant angles with $\cos(\alpha) = \frac{\sqrt{3}}{2}$ and $\sin(\beta) = \frac{\sqrt{6}}{6}$.
Find $\cos(\alpha + \beta)$.
$\cos(\alpha + \beta) = $
No solution
$\frac{\sqrt{3}}{2}\frac{\sqrt{30}}{6} - \frac{\sqrt{1}}{2}\frac{\sqrt{6}}{6} \approx 0.58644526981016$

Let α and β be first quadrant angles with cos(α) = (√(3))/(2) and sin(β) = (√(6))/(6).
Find cos(α + β).
cos(α + β) =
No solution
(√(3))/(2)(√(30))/(6) - (√(1))/(2)(√(6))/(6)≈ 0.58644526981016

Added by Christopher D.

Introductory and Intermediate Algebra for College Students 4th

Robert Blitzer 4th Edition

Instant Answer

Solved by Expert Matthew Wagner

02/21/2024

Step 1

Since $\alpha$ and $\beta$ are in the first quadrant, we can find $\sin(\alpha)$ and $\cos(\beta)$ using the Pythagorean identities. $\sin^2(\alpha) + \cos^2(\alpha) = 1$ $\sin^2(\alpha) + (\frac{\sqrt{3}}{2})^2 = 1$ $\sin^2(\alpha) + \frac{3}{4} = Show more…

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Let $\alpha$ and $\beta$ be first quadrant angles with $\cos(\alpha) = \frac{\sqrt{3}}{2}$ and $\sin(\beta) = \frac{\sqrt{6}}{6}$. Find $\cos(\alpha + \beta)$. $\cos(\alpha + \beta) = \text{________}$ No solution $\frac{\sqrt{3}}{2}\frac{\sqrt{30}}{6} - \frac{\sqrt{1}}{2}\frac{\sqrt{6}}{6} \approx 0.58644526981016$