sin2 + cos2 = 1
So, (1 - 2*cos2)/(sin*cos) = (sin2 + cos2 - 2*cos2)/(sin*cos)
= (sin2 - cos2)/(sin*cos)
= sin2/(sin*cos) - cos2/(sin*cos)
= sin/cos - cos-sin
= tan - cot
opposite over adjacent
y=3cos(x) peroid is 2pie
49 = (±7)2 x2<=49 => -7 <= x <= 7
The Unit Circle is a circle that has a radius of 1 and a center at the origin. If you look at the unit circle 90 degrees is at the point (0,1). Cosine is equal to the x value of a point on the Unit Circle. The line created to the point (0,1) on the unit circle when the degree is 90 is completely vertical, which in turn makes the x value 0 and thus, cosine of 90 = 0. Read more >> Options >> http://www.answers.com?initiator=FFANS
It is impossible but if it were x squared plus 2x minus 15 the equation would be (x+5) (x-3) with x being equal to either -5 or 3. If the original problem was x squared minus 2x minus 15 the equation would be (x-5)(x+3) and x would be equal to either 5 or -3
No, it is not. To be correct, the expression requires parenthesis, which are missing.
Cosine squared theta = 1 + Sine squared theta
No, it does not.
No, they do not.
One plus cosecant squared x is equal to cotangent squared x.
pineapple
You could just pull out the half: it will be (1/2) cos squared x.
Since you didn't specify which trigonometric function you're using, I'll give you all of them.120 in Degreessin120 ~ 0.87cos120 ~ -0.5tan120 ~ -1.73csc120 ~ 1.15sec120 = -2cot120 ~ -0.58Answer in Degreesarctan120 ~ 89.52arccot120 ~ 0.48120 in Radianssin120 ~ 0.58cos120 ~ 0.81tan120 ~ 0.71csc120 ~ 1.72sec120 ~ 1.23cot120 ~ 1.4Answer in Radiansarctan120 ~ 1.56arccot120 ~ 0.008
The tangent of an angle theta is defined as sine(theta) divided by cosine(theta). Since the sine and cosine are Y and X on the unit circle, then tangent(theta) is Y divided by X. The tangent of a function at a point is the line going through that point which has slope equal to the first deriviative of the function at that point.
Tan = o/a Tangent of an angle = opposite over adjacent. Here are the other Trig. functions. SINe(angle) = opposite/hypotenuse COSine(angle) = adjacent/hypotenuse COTangent(angle) = adjacent/opposite Cosecant(CSC)(angle) = hypotenuse/oppositre SECant(angle) = hypotenuse/ adjcent.
2 x cosine squared x -1 which also equals cos (2x)
The tangent function is equal to the sine divided by the cosine. In quadrant III, both sin and cos are negative - and a negative divided by another negative is positive. Thus it follows that the tangent is positive in QIII.