Asked by Mars
Find the value of the six trigonometric functions.
t = 7(pi)/4
Can you show me step by step so I can see how this is solved? Also, do I need to graph this first in order to understand or begin the problem? I am trying to self teach - but having problems.
t = 7(pi)/4
Can you show me step by step so I can see how this is solved? Also, do I need to graph this first in order to understand or begin the problem? I am trying to self teach - but having problems.
Answers
Answered by
Henry
t = 7(pi)/4 Radians.
7(p1)/4 * 180/(pi) = 315 deg,
315 deg. is in 4th quadrant where sine
and tangent are negative.
sin315 = -0.707 = -sqrt(2)/2 = y/r,
x^2 + y^2 = r^2,
x^2 + (-sqrt2)^2 = 2^2,
x^2 + 2 = 4,
x^2 = 2,
x = sqrt(2).
sin315 = y/r = -sqrt(2)/2,
cos315 = x/r = sqrt(2)/2,
tan315 = y/x = -sqrt(2)/sqrt(2) = -1,
csc315 = 1/sin315 = 2/sqrt(2) = 2sqrt2/2 = sqrt2.
sec315 = 1/cos315 = 2/swqrt2 = 2sqrt2/2
= sqrt2.
cot315 = 1/tan315 = 1/-1 = -1.
Notice that no radicals were left in the denominators.
The 6 trig functions in decimal form
are easily found using the calculator:
sin315 = -0.707,
cos315 = 0.707,
tan315 = -1,
csc315 = 1/sin315 = -1.414,
sec315 = 1/cos315 = 1.414,
cot315= 1/tan315 = -1.
7(p1)/4 * 180/(pi) = 315 deg,
315 deg. is in 4th quadrant where sine
and tangent are negative.
sin315 = -0.707 = -sqrt(2)/2 = y/r,
x^2 + y^2 = r^2,
x^2 + (-sqrt2)^2 = 2^2,
x^2 + 2 = 4,
x^2 = 2,
x = sqrt(2).
sin315 = y/r = -sqrt(2)/2,
cos315 = x/r = sqrt(2)/2,
tan315 = y/x = -sqrt(2)/sqrt(2) = -1,
csc315 = 1/sin315 = 2/sqrt(2) = 2sqrt2/2 = sqrt2.
sec315 = 1/cos315 = 2/swqrt2 = 2sqrt2/2
= sqrt2.
cot315 = 1/tan315 = 1/-1 = -1.
Notice that no radicals were left in the denominators.
The 6 trig functions in decimal form
are easily found using the calculator:
sin315 = -0.707,
cos315 = 0.707,
tan315 = -1,
csc315 = 1/sin315 = -1.414,
sec315 = 1/cos315 = 1.414,
cot315= 1/tan315 = -1.
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