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.
Other 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)
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)
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)
cos*cot + sin = cos*cos/sin + sin = cos2/sin + sin = (cos2 + sin2)/sin = 1/sin = cosec
To show that (cos tan = sin) ??? Remember that tan = (sin/cos) When you substitute it for tan, cos tan = cos (sin/cos) = sin QED
'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
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)
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
[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,
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.
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)
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)
(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
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)
cos*cot + sin = cos*cos/sin + sin = cos2/sin + sin = (cos2 + sin2)/sin = 1/sin = cosec
sin cubed + cos cubed (sin + cos)( sin squared - sin.cos + cos squared) (sin + cos)(1 + sin.cos)
e^(i*x)=cos(x)+i*sin(x) pretty sweet formula
To show that (cos tan = sin) ??? Remember that tan = (sin/cos) When you substitute it for tan, cos tan = cos (sin/cos) = sin QED