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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)
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'
1-2sin squared x equals?
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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)
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 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)
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 is sin squared minus 1?
-cos2(x)1 = sin2(x) +cos2(x)1 - cos2(x) = sin2(x)-cos2(x) = sin2(x) - 1
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.
Sine plus cosine equals?
There is no real significance to sine plus cosine, now sin2(x) + cos2(x) = 1 for any x, where sin2(x) means to take the sign of the number, then square that value.
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
Sec²x - tan²x equals?
sec2(x) - tan2(x)= 1/cos2(x) - sin2(x)/cos2(x)= (1 - sin2(x)) / cos2(x)= cos2(x) / cos2(x)= 1
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 can you write an expression for cos x in term of sin x using pythagorean theorem?
cos2 x + sin2 x = 1 cos2 x = 1 - sin2 x
How do you simplify cos times cot plus sin?
cos*cot + sin = cos*cos/sin + sin = cos2/sin + sin = (cos2 + sin2)/sin = 1/sin = cosec
1-2sin squared x equals?
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