Anyone know how to do trig? Given tan Ф = 4/5, find all other trig functions.
1 answer:
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Answer:
- sin(Ф) = (4/41)√41; csc(Ф) = (√41)/4
- cos(Ф) = (5/41)√41; sec(Ф) = (√41)/5
- tan(Ф) = 4/5; cot(Ф) = 5/4
Step-by-step explanation:
The mnemonic SOH CAH TOA can help you with three of the functions. Then you need to remember the relations between those and the other three.
Tan = Opposite/Adjacent
tan(Ф) = 4/5
This suggests you can draw a triangle like the one attached that has adjacent side 5 and opposite side 4. Then hypotenuse can be found from the Pythagorean theorem:
AC² = AB² +BC² = 5² +4² = 41
AC = √41 . . . . . hypotenuse
Then the other trig functions are ...
Sin = Opposite/Hypotenuse
sin(Ф) = 4/√41 = (4/41)√41
Cos = Adjacent/Hypotenuse
cos(Ф) = 5/√41 = (5/41)√41
And the reciprocal functions are ...
csc(Ф) = 1/sin(Ф) = (√41)/4
sec(Ф) = 1/cos(Ф) = (√41)/5
cot(Ф) = 1/tan(Ф) = 5/4
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Answer:
Step-by-step explanation:
The standard form of an equation for a straight line is y=mx+b, where m is the slope and b is the y-intercept (the value of y when x = 0).
We can calculate the slope from the two given points, (6,-3) and (-6,-5). Slope is Rise/Run, where Rise is the change in y and Run is the change in x.
From the two given points, starting at (-6,-5) and going to (6,-3):
Rise = (-3 - (-5)) = +2
Run = (6 - (-6)) = 12
Rise/Run (slope) = 2/12 or 1/6
The equation becomes y = (1/6)x + b
We can find b by enterieng either of the two given points and solving for b. I'll pick (6,-3):
y = (1/6)x + b
-3 = (1/6)*(6) + b
-3 = 1 + b [Now you can see why I chose (6,-3)]
b = -4
The equation is y = (1/6)x - 4
Check this with a DESMOS graph (attached).
