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What is the relationship between the sine of an angle and cosine of the other acute angle in the same triangle?
In a right triangle, the sine of one acute angle is equal to the cosine of the other acute angle. This relationship arises from the definitions of sine and cosine: for an angle ( A ), ( \sin(A) ) is the ratio of the length of the opposite side to the hypotenuse, while ( \cos(B) ), where ( B ) is the other acute angle, is the ratio of the length of the adjacent side to the hypotenuse. Since the two angles are complementary (summing to 90 degrees), this relationship can be expressed as ( \sin(A) = \cos(90^\circ - A) ).
Given that Sin x equals 0.8 and x is acute write down the value of Cos x?
cos2x = 1 - sin2x = 1 - 0.64 = 0.36 So cos x = +/- 0.6 Since x is acute, cos x is +ve, so cos x = 0.6
How do you prove this trigonometric relationship sin3A equals 3sinA cos 2 A - sin 3 A?
sin(3A) = sin(2A + A) = sin(2A)*cos(A) + cos(2A)*sin(A)= sin(A+A)*cos(A) + cos(A+A)*sin(A) = 2*sin(A)*cos(A)*cos(A) + {cos^2(A) - sin^2(A)}*sin(A) = 2*sin(A)*cos^2(A) + sin(a)*cos^2(A) - sin^3(A) = 3*sin(A)*cos^2(A) - sin^3(A)
How do you verify the identity of cos θ tan θ equals sin θ?
To show that (cos tan = sin) ??? Remember that tan = (sin/cos) When you substitute it for tan, cos tan = cos (sin/cos) = sin QED
Sin 15 plus cos 105 equals?
Sin 15 + cos 105 = -1.9045
If θ is an acute angle and sin θ equals 2 then cos θ?
it equals 4
X is acute and sin x equals cos x Find cos?
x = 45 degrees sin(x) = cos(x) = 1/2 sqrt(2)
How do you calculate trigonometric ratios of obtuse angles?
If ø is an obtuse angle then (180 - ø) is an acute angle and: sin ø = sin (180 - ø) cos ø = -cos (180 - ø) tan ø = -tan (180 - ø)
What is the relationship between the sine of an angle and cosine of the other acute angle in the same triangle?
In a right triangle, the sine of one acute angle is equal to the cosine of the other acute angle. This relationship arises from the definitions of sine and cosine: for an angle ( A ), ( \sin(A) ) is the ratio of the length of the opposite side to the hypotenuse, while ( \cos(B) ), where ( B ) is the other acute angle, is the ratio of the length of the adjacent side to the hypotenuse. Since the two angles are complementary (summing to 90 degrees), this relationship can be expressed as ( \sin(A) = \cos(90^\circ - A) ).
What is cos2 A?
Cos(2A) = Cos(A + A) Double Angle Indentity Cos(A+A) = Cos(A)Cos(A) - Sin(A)Sin(A) => Cos^(2)[A] - SIn^(2)[A] => Cos^(2)[A] - (1 - Cos^(2)[A] => 2Cos^(2)[A] - 1
Write down the acute angle whose sine is equal to Cos 70?
cos70=sin20 so angle is 20, because cosA=sin(90-A)
What is this expression as the cosine of an angle cos30cos55 plus sin30sin55?
cos(30)cos(55)+sin(30)sin(55)=cos(30-55) = cos(-25)=cos(25) Note: cos(a)=cos(-a) for any angle 'a'. cos(a)cos(b)+sin(a)sin(b)=cos(a-b) for any 'a' and 'b'.
Given that Sin x equals 0.8 and x is acute write down the value of Cos x?
cos2x = 1 - sin2x = 1 - 0.64 = 0.36 So cos x = +/- 0.6 Since x is acute, cos x is +ve, so cos x = 0.6
How do you prove this trigonometric relationship sin3A equals 3sinA cos 2 A - sin 3 A?
sin(3A) = sin(2A + A) = sin(2A)*cos(A) + cos(2A)*sin(A)= sin(A+A)*cos(A) + cos(A+A)*sin(A) = 2*sin(A)*cos(A)*cos(A) + {cos^2(A) - sin^2(A)}*sin(A) = 2*sin(A)*cos^2(A) + sin(a)*cos^2(A) - sin^3(A) = 3*sin(A)*cos^2(A) - sin^3(A)
How do you verify the identity of cos θ tan θ equals sin θ?
To show that (cos tan = sin) ??? Remember that tan = (sin/cos) When you substitute it for tan, cos tan = cos (sin/cos) = sin QED
How would you prove left cosA plus sinA right times left cos2A plus sin2A right equals cosA plus sin3A?
You need to make use of the formulae for sin(A+B) and cos(A+B), and that cos is an even function: sin(A+B) = cos A sin B + sin A cos B cos(A+B) = cos A cos B - sin A sin B cos even fn → cos(-x) = cos(x) To prove: (cos A + sin A)(cos 2A + sin 2A) = cos A + sin 3A The steps are to work with the left hand side, expand the brackets, collect [useful] terms together, apply A+B formula above (backwards) and apply even nature of cos function: (cos A + sin A)(cos 2A + sin 2A) = cos A cos 2A + cos A sin 2A + sin A cos 2A + sin A sin 2A = (cos A cos 2A + sin A sin 2A) + (cos A sin 2A + sin A cos 2A) = cos(A - 2A) + sin(A + 2A) = cos(-A) + sin 3A = cos A + sin 3A which is the right hand side as required.
Verify that sin minus cos plus 1 divided by sin plus cos subtract 1 equals sin plus 1 divided by cos?
[sin - cos + 1]/[sin + cos - 1] = [sin + 1]/cosiff [sin - cos + 1]*cos = [sin + 1]*[sin + cos - 1]iff sin*cos - cos^2 + cos = sin^2 + sin*cos - sin + sin + cos - 1iff -cos^2 = sin^2 - 11 = sin^2 + cos^2, which is true,
