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The answer depends on what the triangle is and what information you have. The formulae which are available for right angled triangles are not applicable for triangles which do not have a right angle. Trigonometric ratios are defined for angles, whether or not the shape is a triangle.
In general, there is no simple method for working out these ratios without a calculator (or reference tables).
If x is the angle, measured in radians, then
sin(x) = x - x^3/3! + x^5/5! - x^7/7! ...
cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! ...
and
tan(x) = sin(x)/cos(x).
But these are hardly calculable without a calculator!
Note that n! = 1*2*3* ... *n for positive integer n.
Wiki User
∙ 8y ago
Wiki User
∙ 8y agoOther calculators can calculate those as well. For example, the scientific calculator on Windows.Since the actual calculations are quite complicated, you should use a calculator to calculate these.
However, just in case you are curious, the calculations work with infinite sums, as below. The idea is to add more and more terms for the infinite sum, until you notice that the terms become so small that your answer is accurate enough.
First, convert the angle to radians. Then, use one of the following formulae (using "^" for power, and "!" for the factorial function):
sin(x) = x - x^3/3! + x^5/5! - x^7/7! ...
cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! ...
For tan(x), a more complicated infinite sum exists. Or just use the identity:
tan(x) = sin(x) / cos(x)
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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 I find the product z1z2 if z1 5(cos20 plus isin20) and z2 8(cos15 plus isin15)?
Like normal expansion of brackets, along with: cos(A + B) = cos A cos B - sin A sin B sin(A + B) = sin A cos B + cos A sin B 5(cos 20 + i sin 20) × 8(cos 15 + i sin 15) = 5×8 × (cos 20 + i sin 20)(cos 15 + i sin 15) = 40(cos 20 cos 15 + i sin 15 cos 20 + i cos 15 sin 20 + i² sin 20 sin 15) = 40(cos 20 cos 15 - sin 20 cos 15 + i(sin 15 cos 20 + cos 15 sin 20)) = 40(cos(20 +15) + i sin(15 + 20)) = 40(cos 35 + i sin 35)
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
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 do you simplify cos theta times csc theta divided by tan theta?
'csc' = 1/sin'tan' = sin/cosSo it must follow that(cos) (csc) / (tan) = (cos) (1/sin)/(sin/cos) = (cos) (1/sin) (cos/sin) = (cos/sin)2
How do you proof the formula sin2A equals 2sinAcosA?
First, note that sin(a+b)=sin(a)cos(b)+sin(b)cos(a)[For a proof, see: www.mathsroom.co.uk/downloads/Compound_Angle_Proof.pptFor the case of b=a, we have:sin (a+a)=sin(a)cos(a)+sin(a)cos(a)sin (2a)=2*sin(a)cos(a)
How do you use the Euler's formula to obtain the sin3X in terms of cosX and sinX?
Using Euler's Formula, you use (cos(x) + i sin(x))^n = cos (nx) + i sin(nx) Now you let n=3 (cos(x) + i sin (x))3 = cos(3x) + i sin (3x) (cos(x))3 + 3(cos(x))2 * i sin(x) + 3cos(x) * i2 (sin(x))3 = cos(3x)+ i sin(3x) (cos(x))3 + i(3sin(x)(cos (x))2) - 3cos(x)(sin(x)2) - i(sin(x))3 = cos (3x) + i sin(3x) Now only use the terms with i in them to figure out what sin(3x) is... 3sin(x)(cos(x))2 - (sin(x))3 = sin(3x) Hope this helps! :D
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,
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.
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)
Solution for tan x is equal to cos x?
if tan x = cos x then sin x / cos x = cos x => sin x = cos x cos x => sin x = cos2 x => sin x = 1 - sin2x => sin2x + sin x - 1 = 0 Using the quadratic formula => 1. sin x = 0.61803398874989484820458683436564 => x = sin-1 (0.61803398874989484820458683436564) or => 2. sin x = -1.6180339887498948482045868343656 => x = sin-1 (-1.6180339887498948482045868343656)
How do you show that 2 sin squared x minus 1 divided by sin x minus cos x equals sin x plus cos x?
(2 sin^2 x - 1)/(sin x - cos x) = sin x + cos x (sin^2 x + sin^2 x - 1)/(sin x - cos x) =? sin x + cos x [sin^2 x - (1 - sin^2 x)]/(sin x - cos x) =? sin x + cos x (sin^2 x - cos^2 x)/(sin x - cos x) =? sin x + cos x [(sin x - cos x)(sin x + cos x)]/(sin x - cos x) =? sin x + cos x sin x + cos x = sin x + cos x
How do I find the product z1z2 if z1 5(cos20 plus isin20) and z2 8(cos15 plus isin15)?
Like normal expansion of brackets, along with: cos(A + B) = cos A cos B - sin A sin B sin(A + B) = sin A cos B + cos A sin B 5(cos 20 + i sin 20) × 8(cos 15 + i sin 15) = 5×8 × (cos 20 + i sin 20)(cos 15 + i sin 15) = 40(cos 20 cos 15 + i sin 15 cos 20 + i cos 15 sin 20 + i² sin 20 sin 15) = 40(cos 20 cos 15 - sin 20 cos 15 + i(sin 15 cos 20 + cos 15 sin 20)) = 40(cos(20 +15) + i sin(15 + 20)) = 40(cos 35 + i sin 35)
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
Factor sin cubed plus cos cubed?
sin cubed + cos cubed (sin + cos)( sin squared - sin.cos + cos squared) (sin + cos)(1 + sin.cos)
What is Euler's calculus formula?
e^(i*x)=cos(x)+i*sin(x) pretty sweet formula
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
