find all solutions t between 360 and 720 degrees, inclusive:
(a) cos t = sin t
(b) ta t = –4.3315
(c) sin t = –0.9397
The Correct Answer and Explanation is :
Let’s solve each part step by step.
(a) ( \cos t = \sin t )
- Set up the equation:
[
\cos t = \sin t
]
Divide both sides by ( \cos t ) (assuming ( \cos t \neq 0 )):
[
1 = \tan t
]
So, we need to find where ( \tan t = 1 ). - Solve for ( t ):
The general solution for ( \tan t = 1 ) is:
[
t = 45^\circ + 180^\circ k
]
where ( k ) is an integer. - Find all solutions between 360° and 720°:
For ( k = 2 ):
[
t = 45^\circ + 180^\circ \times 2 = 405^\circ
]
For ( k = 3 ):
[
t = 45^\circ + 180^\circ \times 3 = 585^\circ
]
So the solutions for this part are:
[
t = 405^\circ, 585^\circ
]
(b) ( \tan t = -4.3315 )
- Use inverse tangent:
[
t = \tan^{-1}(-4.3315)
]
Using a calculator:
[
t \approx -77.1^\circ
] - Find all solutions between 360° and 720°:
Since the tangent function has a period of 180°, add 180° repeatedly to find solutions within the given range. For the first solution, add 360°:
[
t = -77.1^\circ + 360^\circ \approx 282.9^\circ
]
Add 180° to move it into the desired range:
[
t = 282.9^\circ + 180^\circ \approx 462.9^\circ
]
Add another 180°:
[
t = 462.9^\circ + 180^\circ \approx 642.9^\circ
]
So, the solutions are:
[
t = 462.9^\circ, 642.9^\circ
]
(c) ( \sin t = -0.9397 )
- Use inverse sine:
[
t = \sin^{-1}(-0.9397)
]
Using a calculator:
[
t \approx -70^\circ
] - Find all solutions between 360° and 720°:
The sine function is negative in the third and fourth quadrants, so the general solution is:
[
t = 360^\circ + 70^\circ = 430^\circ
]
and
[
t = 360^\circ + 180^\circ – 70^\circ = 650^\circ
]
So the solutions are:
[
t = 430^\circ, 650^\circ
]
Summary of solutions:
- (a) ( t = 405^\circ, 585^\circ )
- (b) ( t = 462.9^\circ, 642.9^\circ )
- (c) ( t = 430^\circ, 650^\circ )
Explanation (300 words):
Each trigonometric equation can be solved by recognizing the relevant identities and using inverse trigonometric functions. In (a), ( \cos t = \sin t ) simplifies to ( \tan t = 1 ), and since tangent has a period of 180°, solutions are found by adding multiples of 180° to the reference angle of 45°. Thus, the solutions between 360° and 720° are 405° and 585°.
In (b), we solve ( \tan t = -4.3315 ) by applying the inverse tangent function. The initial angle is approximately -77.1°, which can be adjusted by adding multiples of 180° to find solutions in the specified interval. This yields solutions at 462.9° and 642.9°.
In (c), ( \sin t = -0.9397 ) leads to ( t = \sin^{-1}(-0.9397) ), which gives a reference angle of approximately -70°. Since sine is negative in the third and fourth quadrants, we add 360° and appropriate multiples of 180° to find the final answers, resulting in solutions at 430° and 650°.
By understanding periodicity and quadrant behavior, we can systematically find all solutions within the desired range.