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cos2A = cos^2A-sin^2A (2A means the double angle .. :|)
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How does 1-sin squared x times 1 plus tan suared x equals 1?
Use these identities: sin2(x) + cos2(x) = 1, and tan(x) = sin(x)/cos(x) For clarity, the functions are written here without their arguments (the "of x" part). (1 - sin2) = cos2 (1 + tan2) = (1 + sin2/cos2) = (cos2+sin2) / cos2 = 1/cos2 Multiply them: (cos2) times (1/cos2) = 1'QED'
What does negative sine squared plus cosine squared equal?
To determine what negative sine squared plus cosine squared is equal to, start with the primary trigonometric identity, which is based on the pythagorean theorem...sin2(theta) + cos2(theta) = 1... and then solve for the question...cos2(theta) = 1 - sin2(theta)2 cos2(theta) = 1 - sin2(theta) + cos2(theta)2 cos2(theta) - 1 = - sin2(theta) + cos2(theta)
Show that cos3t equals 4cos cubed t - 3cos t?
cos(3t) = cos(2t + t) = cos(2t)*cos(t) - sin(2t)*sin(t) = [cos2(t) - sin2(t)]*cos(t) - 2*cos(t)*sin(t)*sin(t) = [cos2(t) - sin2(t)]*cos(t) - 2*cos(t)*sin2(t) then, since sin2(t) = 1 - cos2(t) = [2*cos2(t) - 1]*cos(t) - 2*cos(t)*[1 - cos2(t)] = 2*cos3(t) - cos(t) - 2*cos(t) + 2*cos3(t) = 4*cos3(t) - 3*cos(t)
1-2sin squared x equals?
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How do you figure out the Trigonometry theorem?
If, by trigonometry theorem you mean the "fundamental theorem of trigonometry," sin2(x) + cos2(x) = 1, it is actually the Pythagorean Theorem. if you have a right triangle with a hypotenuse of one, sin(x) is one leg, and cos(x) is the other. The Pythagorean Theorem states that a2 + b2 = c2 and therefore sin2(x) + cos2(x) = 1.
What is the solution to cos2 plus cos2tan2 equals 1?
cos2 + cos2tan2 = cos2 + cos2*sin2/cos2 = cos2 + sin2 which is identically equal to 1. So the solution is all angles.
Why does cos2 equals pi?
Cos2 doesn't equal pi; Cos2 equals roughly -0.416 (with radians).
Sec²x - tan²x equals?
sec2(x) - tan2(x)= 1/cos2(x) - sin2(x)/cos2(x)= (1 - sin2(x)) / cos2(x)= cos2(x) / cos2(x)= 1
What is sin squared minus 1?
-cos2(x)1 = sin2(x) +cos2(x)1 - cos2(x) = sin2(x)-cos2(x) = sin2(x) - 1
How does 1-sin squared x times 1 plus tan suared x equals 1?
Use these identities: sin2(x) + cos2(x) = 1, and tan(x) = sin(x)/cos(x) For clarity, the functions are written here without their arguments (the "of x" part). (1 - sin2) = cos2 (1 + tan2) = (1 + sin2/cos2) = (cos2+sin2) / cos2 = 1/cos2 Multiply them: (cos2) times (1/cos2) = 1'QED'
How can you prove that 1-2 cosine squared over sine times cosine is equal to tangent minus cotangent?
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
What does negative sine squared plus cosine squared equal?
To determine what negative sine squared plus cosine squared is equal to, start with the primary trigonometric identity, which is based on the pythagorean theorem...sin2(theta) + cos2(theta) = 1... and then solve for the question...cos2(theta) = 1 - sin2(theta)2 cos2(theta) = 1 - sin2(theta) + cos2(theta)2 cos2(theta) - 1 = - sin2(theta) + cos2(theta)
In maths does one minus two sine squared an equal cos squared a minus sine squared a 1-2sin2an equals sin2a-cos2a?
No.Remember: sin2 + cos2 = 1So, in place of (1 - 2 sin2) we can write (sin2 + cos2 - 2 sin2).Massage that around slightly: (sin2 + cos2 - 2 sin2) = cos2 - sin2That's not equal to (sin2 - cos2), which is the original question we were asked.
How would you solve and show work for cos2 theta if cos squared theta equals 1 and theta is in the 4th quadrant?
cos2(theta) = 1 cos2(theta) + sin2(theta) = 1 so sin2(theta) = 0 cos(2*theta) = cos2(theta) - sin2(theta) = 1 - 0 = 1
How does sec-squared x equals 1 plus tan-squared x?
Let s = sin x; c = cos x. By definition, sec x = 1/cos x = 1/c; and tan x = (sin x) / (cos x) = s/c. We know, also, that s2 + c2 = 1. Then, dividing through by c2, we have, (s2/c2) + 1 = (1/c2), or (s/c)2 + 1 = (1/c)2; in other words, we have, tan2 x + 1 = sec2 x.
How do you simplify tan2 x-1 over 1-sec2 x?
tan2 x + 1 = sec2 x ⇒ 1 - sec2 x = -tan2 x ⇒ (tan2 x - 1) / (1 - sec2 x) = (tan2 x - 1) / -tan2 x = (tan2 x) / (-tan2 x) - 1 / (-tan2 x) = -1 + cot2 x = cot2 x - 1 If you cannot remember tan2 x + 1 = sec2 x, remember and start from: sin2 x + cos2 x = 1 (which is used often) and divide each side by cos2 x: (sin2 x + cos2 x) / cos2 x = 1 / cos2 x ⇒ sin2 x / cos2 x + cos2 x / cos2 x = 1 / cos2 x But sin x / cos x = tan x; and 1/cos x = sec x ⇒ tan2 x + 1 = sec2 x Also, cot x = 1/tan x
Where is cos2 on a ti-83 plus calculator?
There is not cos2 button on a TI-83 plus. You will need to enter the cosine function and then square it. (Press the x2 button to get the squared function.) To type cos2(90) on a TI-83 plus, for example, type: cos(90)2
