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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?
-Sin^(2)(Theta) + Cos^(2)Theta => Cos^(2)Theta - Sin^(2)Theta Factor (Cos(Theta) - Sin(Theta))( Cos(Theta) + Sin(Theta)) #Is the Pythagorean factors . Or -Sin^(2)Theta = -(1 - Cos^(2)Theta) = Cos(2)Theta - 1 Substitute Cos^(2)Thetqa - 1 + Cos^(2) Theta = 2Cos^(2)Theta - 1
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.
What is the value of sinX wrt cosX?
sin(x) = sqrt[ 1 - cos2(x) ]
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
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
What is the number that has the same absolute value as 23?
-23 The absolute value of -23 and 23 is 23.
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
