1. Value of cos2theta when tantheta = 3/4 and pi < theta < 3pi/2
2. Solve 5cos^2x + 4 cos x = 1 if 0 <= x <= 2pi
+9491 1. Find the value of cos( 2θ ) when tan( θ ) = 3/4 and π < θ < 3π/2
To use the double-angle formula for cos, let's find sin( θ ) and cos( θ ) with a sketch and the Pythagorean theorem.
sin( θ ) = opposite / hypotenuse = -3/5
cos( θ ) = adjacent / hypotenuse = -4/5
| cos( 2θ ) = cos2( θ ) - sin2( θ )____ | by the double-angle formula for cos |
| cos( 2θ ) = ( -4/5 )2 - ( -3/5 )2 | by subsituting ( -4/5 ) in for cos( θ ) and ( -3/5 ) in for sin( θ ) |
| cos( 2θ ) = 16/25 - 9/25 | |
| cos( 2θ ) = 7/25 |
+9491 2. Solve 5cos2(x) + 4cos(x) = 1 if 0 ≤ x ≤ 2π
| 5cos2(x) + 4cos(x) = 1 | ____ | |||||
| Subtract 1 from both sides of the equation. | ||||||
| 5cos2(x) + 4cos(x) - 1 = 0 | Notice that this is a quadratic equation. | |||||
| Split 4cos(x) into two terms so that the product of their coefficients is -5 | ||||||
| 5cos2(x) + 5cos(x) - 1cos(x) - 1 = 0 | ||||||
| Factor 5cos(x) out of first two terms, factor -1 out of last two terms. | ||||||
| 5cos(x)(cos(x) + 1) - 1(cos(x) + 1) = 0 | ||||||
| Factor (cos(x) + 1) out of both remaining terms. | ||||||
| ( cos(x) + 1 )( 5cos(x) - 1 ) = 0 | ||||||
| Set each factor equal to zero. | ||||||
| cos(x) + 1 = 0 | __or__ | 5cos(x) - 1 = 0 |
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| cos(x) = -1 | cos(x) = 1/5 |
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| x = π | x ≈ 1.369 x ≈ 4.914 | Only use values of x in the interval given |
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