Trigonometry Examples

Solve for θ in Degrees cos(theta/2)=-( square root of 2)/2
cos(θ2)=-22cos(θ2)=22
Step 1
Take the inverse cosine of both sides of the equation to extract θθ from inside the cosine.
θ2=arccos(-22)θ2=arccos(22)
Step 2
Simplify the right side.
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Step 2.1
The exact value of arccos(-22)arccos(22) is 135135.
θ2=135θ2=135
θ2=135θ2=135
Step 3
Multiply both sides of the equation by 22.
2θ2=21352θ2=2135
Step 4
Simplify both sides of the equation.
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Step 4.1
Simplify the left side.
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Step 4.1.1
Cancel the common factor of 22.
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Step 4.1.1.1
Cancel the common factor.
2θ2=2135
Step 4.1.1.2
Rewrite the expression.
θ=2135
θ=2135
θ=2135
Step 4.2
Simplify the right side.
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Step 4.2.1
Multiply 2 by 135.
θ=270
θ=270
θ=270
Step 5
The cosine function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from 360 to find the solution in the third quadrant.
θ2=360-135
Step 6
Solve for θ.
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Step 6.1
Multiply both sides of the equation by 2.
2θ2=2(360-135)
Step 6.2
Simplify both sides of the equation.
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Step 6.2.1
Simplify the left side.
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Step 6.2.1.1
Cancel the common factor of 2.
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Step 6.2.1.1.1
Cancel the common factor.
2θ2=2(360-135)
Step 6.2.1.1.2
Rewrite the expression.
θ=2(360-135)
θ=2(360-135)
θ=2(360-135)
Step 6.2.2
Simplify the right side.
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Step 6.2.2.1
Simplify 2(360-135).
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Step 6.2.2.1.1
Subtract 135 from 360.
θ=2225
Step 6.2.2.1.2
Multiply 2 by 225.
θ=450
θ=450
θ=450
θ=450
θ=450
Step 7
Find the period of cos(θ2).
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Step 7.1
The period of the function can be calculated using 360|b|.
360|b|
Step 7.2
Replace b with 12 in the formula for period.
360|12|
Step 7.3
12 is approximately 0.5 which is positive so remove the absolute value
36012
Step 7.4
Multiply the numerator by the reciprocal of the denominator.
3602
Step 7.5
Multiply 360 by 2.
720
720
Step 8
The period of the cos(θ2) function is 720 so values will repeat every 720 degrees in both directions.
θ=270+720n,450+720n, for any integer n
 [x2  12  π  xdx ]