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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The values of k for the different quadratic equation solutions are as follows
a the equation 2x² - x + 3k = 0 has two distinct real roots
- k < 1/24
b. the equation 5x² - 2x + (2k − 1) = 0 has equal roots
- k = 3/5
ci the equation -x² + 3x + (k + 1) = 0 has real roots
- k > -3.25
d the equation 3kx² - 3x + 2 = 0 has no real solutions
- k < ± 1.633
Quadratic equations of the form ax² + bx + c = 0 is solved using the formula
OR
The equation b² - 4ac is called the discriminant and it is used as follows
To solve the equation and get two real roots: 2x² - x + 3k = 0
- b² - 4ac > 0
substituting the values gives
(-1)² - 4 * 2 * 3k > 0
1 - 24k > 0
1 > 24k
divide through by coefficient of k
k < 1/24
To solve the equation and get equal roots: 5x² - 2x + (2k − 1) = 0
- b² - 4ac = 0
substituting the values gives
(-2)² - 4 * 5 * (2k - 1) = 0
4 - 40k + 20 = 0
-40k = -24
divide through by coefficient of k
k = 3/5
To solve the equation and get real roots -x² + 3x + (k + 1) = 0
- b² - 4ac > 0
substituting the values gives
(3)² - 4 * -1 * (k+1) > 0
9 + 4k + 4> 0
4k > -13
divide through by coefficient of k
k > -3.25
To solve the equation and get no real solutions 3kx² - 3x + 2 = 0
- b² - 4ac < 0
substituting the values gives
(-3)² - 4 * 3k * 2 < 0
9 - 24k² > 0
9 > 24k²
divide through by coefficient of k²
k² < 24/9
k < ± 1.633
Learn more about roots of quadratic equations: brainly.com/question/26926523
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