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Given that tan x=(1)/(sqrt3) and sin y=(sqrt28)/(6), and that angles x and y are both in Quadrant I, find the exact value of cos(x+y), in simplest radical form. Answer:

Given that tan⁡x=13 \tan x=\frac{1}{\sqrt{3}} and sin⁡y=286 \sin y=\frac{\sqrt{28}}{6} , and that angles x x and y y are both in Quadrant I, find the exact value of cos⁡(x+y) \cos (x+y) , in simplest radical form.\newlineAnswer:

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Q. Given that tan⁡x=13 \tan x=\frac{1}{\sqrt{3}} and sin⁡y=286 \sin y=\frac{\sqrt{28}}{6} , and that angles x x and y y are both in Quadrant I, find the exact value of cos⁡(x+y) \cos (x+y) , in simplest radical form.\newlineAnswer:
  1. Find tan⁡x\tan x: We know that tan⁡x=13\tan x = \frac{1}{\sqrt{3}}. Since xx is in Quadrant I, both sine and cosine are positive. We can find cos⁡x\cos x using the Pythagorean identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1, and the fact that tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x}.
  2. Express sin⁡x\sin x: First, let's express sin⁡x\sin x in terms of tan⁡x\tan x. We have tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x}, so sin⁡x=tan⁡x⋅cos⁡x\sin x = \tan x \cdot \cos x. We know tan⁡x=13\tan x = \frac{1}{\sqrt{3}}, so sin⁡x=(13)⋅cos⁡x\sin x = \left(\frac{1}{\sqrt{3}}\right) \cdot \cos x.
  3. Use Pythagorean identity: Now, we use the Pythagorean identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1. Substituting sin⁡x\sin x with (1/3)⋅cos⁡x(1/\sqrt{3}) \cdot \cos x, we get ((1/3)⋅cos⁡x)2+cos⁡2x=1((1/\sqrt{3}) \cdot \cos x)^2 + \cos^2 x = 1.
  4. Solve for cos⁡x\cos x: Simplifying the equation, we have (1/3)⋅cos⁡2x+cos⁡2x=1(1/3) \cdot \cos^2 x + \cos^2 x = 1. Combining like terms, we get (1/3+1)⋅cos⁡2x=1(1/3 + 1) \cdot \cos^2 x = 1, which simplifies to (4/3)⋅cos⁡2x=1(4/3) \cdot \cos^2 x = 1.
  5. Find cos⁡y\cos y: To find cos⁡x\cos x, we solve for cos⁡2x\cos^2 x by multiplying both sides by 34\frac{3}{4}, which gives us cos⁡2x=(34)\cos^2 x = \left(\frac{3}{4}\right). Taking the square root of both sides, we get cos⁡x=34=32\cos x = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}, since xx is in Quadrant I and cosine is positive.
  6. Substitute sin⁡y\sin y: Next, we need to find cos⁡y\cos y. We are given sin⁡y=28/6\sin y = \sqrt{28}/6. Using the Pythagorean identity sin⁡2y+cos⁡2y=1\sin^2 y + \cos^2 y = 1, we can find cos⁡y\cos y.
  7. Simplify the equation: Substitute sin⁡y\sin y with 28/6\sqrt{28}/6 into the identity: (28/6)2+cos⁡2y=1(\sqrt{28}/6)^2 + \cos^2 y = 1. Simplifying, we get (28/36)+cos⁡2y=1(28/36) + \cos^2 y = 1.
  8. Find cos⁡y\cos y: Subtracting 2836\frac{28}{36} from both sides, we get cos⁡2y=1−2836\cos^2 y = 1 - \frac{28}{36}. Simplifying further, we get cos⁡2y=3636−2836=836\cos^2 y = \frac{36}{36} - \frac{28}{36} = \frac{8}{36}.
  9. Use cosine sum formula: Taking the square root of both sides to find cos⁡y\cos y, we get cos⁡y=836=86\cos y = \sqrt{\frac{8}{36}} = \frac{\sqrt{8}}{6}. Since yy is in Quadrant I, cos⁡y\cos y is positive, so cos⁡y=86=2∗46=2∗26=23\cos y = \frac{\sqrt{8}}{6} = \frac{\sqrt{2*4}}{6} = \frac{2*\sqrt{2}}{6} = \frac{\sqrt{2}}{3}.
  10. Find sin⁡x\sin x: Now we have both cos⁡x=3/2\cos x = \sqrt{3}/2 and cos⁡y=2/3\cos y = \sqrt{2}/3. We can use the cosine sum formula to find cos⁡(x+y)\cos(x+y): cos⁡(x+y)=cos⁡x⋅cos⁡y−sin⁡x⋅sin⁡y\cos(x+y) = \cos x \cdot \cos y - \sin x \cdot \sin y.
  11. Substitute all values: We already have cos⁡x\cos x and cos⁡y\cos y. We need to find sin⁡x\sin x, which we can get from tan⁡x=13\tan x = \frac{1}{\sqrt{3}} and cos⁡x=32\cos x = \frac{\sqrt{3}}{2}. Since tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x}, we have sin⁡x=tan⁡x⋅cos⁡x=(13)⋅(32)=12\sin x = \tan x \cdot \cos x = \left(\frac{1}{\sqrt{3}}\right) \cdot \left(\frac{\sqrt{3}}{2}\right) = \frac{1}{2}.
  12. Simplify the expression: We already have sin⁡y=286\sin y = \frac{\sqrt{28}}{6}. Now we can substitute all values into the cosine sum formula: cos⁡(x+y)=(32)⋅(23)−(12)⋅(286)\cos(x+y) = \left(\frac{\sqrt{3}}{2}\right) \cdot \left(\frac{\sqrt{2}}{3}\right) - \left(\frac{1}{2}\right) \cdot \left(\frac{\sqrt{28}}{6}\right).
  13. Combine the terms: Simplifying the expression, we get cos⁡(x+y)=(3⋅2)/(2⋅3)−(1⋅28)/(2⋅6)=(6/6)−(28/12)\cos(x+y) = (\sqrt{3}\cdot\sqrt{2})/(2\cdot3) - (1\cdot\sqrt{28})/(2\cdot6) = (\sqrt{6}/6) - (\sqrt{28}/12).
  14. Simplify the numerator: To combine the terms, we need a common denominator. Multiplying the first term by 22\frac{2}{2}, we get 2612\frac{2\sqrt{6}}{12} - 2812\frac{\sqrt{28}}{12} = 26−2812\frac{2\sqrt{6} - \sqrt{28}}{12}.
  15. Final simplification: Simplifying the numerator, we get (26−4×7)/12=(26−27)/12(2\sqrt{6} - \sqrt{4\times7})/12 = (2\sqrt{6} - 2\sqrt{7})/12.
  16. Final simplification: Simplifying the numerator, we get (26−4×7)/12=(26−27)/12(2\sqrt{6} - \sqrt{4\times7})/12 = (2\sqrt{6} - 2\sqrt{7})/12.Finally, we can simplify the expression by dividing both numerator and denominator by 22, which gives us (6−7)/6(\sqrt{6} - \sqrt{7})/6 as the exact value of cos⁡(x+y)\cos(x+y) in simplest radical form.

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