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Find sinx+cosx if sinx*cosx=3/8

User Rewolf
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2 Answers

7 votes

Final answer:

To find sinx + cosx, we use trigonometric identities to rewrite the equation sin(x) × cos(x) = 3/8. By substituting known identities, we find that sin(x) + cos(x) = (3√(2) + √(14)) / 8.

Step-by-step explanation:

Double-angle formula for sine: We can use the double-angle formula for sine, which states: sin(2x) = 2 × sin(x) × cos(x).

Rearrange for sin(x) × cos(x): Rearranging the formula for sin(x) × cos(x), we get: sin(x) × os(x) = sin(2x) / 2.

Substitute the given product: Substituting the given product of sin(x) × cos(x) = 3/8, we get: sin(2x) = 2 × (3/8) = 3/4.

Find sin(x) and cos(x) using the Pythagorean identity: Now, we can use the Pythagorean identity for sine and cosine, which states: sin²(x) + cos²(x) = 1.

Rearrange for cos²(x): Rearranging the equation for cos²(x), we get: cos²(x) = 1 - sin²(x).

Substitute the double-angle formula for sin²(x): Since we know sin(2x) = 3/4, we can find sin²(x) using the half-angle formula: sin²(x) = (1 - cos(2x)) / 2. Substituting for sin(2x), we get: sin²(x) = (1 - 3/4) / 2 = 1/8.

Find cos²(x): Substituting the calculated sin²(x) back into the equation for cos^2(x), we get: cos²(x) = 1 - 1/8 = 7/8.

Take the square root to find cos(x): Since cos(x) can be positive or negative, we need to take the square root of both sides to find two possible values: cos(x) = ±√(7/8) = ±√(7) / 2√(2).

Find sin(x) + cos(x) with the same sign: Since sin(x) × cos(x) is positive, both sin(x) and cos(x) must have the same sign. Therefore, we choose the positive value for cos(x): cos(x) = √(7) / 2√(2).

Substitute back into the double-angle formula: Using the positive value of cos(x), we can find sin(x) from the double-angle formula: sin(x) = 3 / (4 × √(2)) = 3√(2) / 8.

Finally, add sin(x) and cos(x): sin(x) + cos(x) = 3√(2) / 8 + √(7) / 2√(2) = (3√(2) + √(14)) / 8.

Therefore, sin(x) + cos(x) = (3√(2) + √(14)) / 8.

User Alex Wih
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9 votes

Answer:


^+_-(√(7))/(2).

Step-by-step explanation:


1.| \ sinx+cosx| =√((sinx+cosx)^2)=√(sin^2x+2sinxcosx+cos^2x)=√(1+2sinxcosx);

2) according to the item 1:


sinx+cosx=^+_-\sqrt{1+2*(3)/(8)} =^+_-\sqrt{(7)/(4)}=^+_-(√(7) )/(2).

User Missy
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4.6k points