Problem:
Solution:
$ \cot(\theta - \frac{\pi}{2}) $
By definition of cotangent, we have
$ = \frac{\cos(\theta - \frac{\pi}{2})}{\sin(\theta - \frac{\pi}{2})} $
By phase shift formulas, we have
$ = \frac{\sin(\theta)}{-\cos(\theta)} $
By definition of tangent, we finally get
$ = -\tan(\theta) $
Monday, January 25, 2016
Thursday, January 21, 2016
Trigonometry problem
Problem:
Solution;
$ \frac{\sec \theta}{\csc \theta - \cot \theta} - \frac{\sec \theta}{\csc \theta + \cot \theta} $
Multiply $ \sin \theta \cos \theta $ above and below:
$ \frac{\sin \theta}{\cos \theta - \cos^2 \theta} - \frac{\sin \theta}{\cos \theta + \cos^2 \theta} $
Make common denominator gives:
$ \frac{\sin \theta(\cos \theta + \cos^2 \theta)}{(\cos \theta + \cos^2 \theta)(\cos \theta - \cos^2 \theta)} - \frac{\sin \theta(\cos \theta - \cos^2 \theta)}{(\cos \theta + \cos^2 \theta)(\cos \theta - \cos^2 \theta)} $
Now do the subtraction
$ \frac{\sin \theta(\cos \theta + \cos^2 \theta)-\sin \theta(\cos \theta - \cos^2 \theta)}{(\cos \theta + \cos^2 \theta)(\cos \theta - \cos^2 \theta)} $
Simplify the numerator by just expand and subtract:
$ \frac{2\sin \theta\cos^2 \theta}{(\cos \theta + \cos^2 \theta)(\cos \theta - \cos^2 \theta)} $
Simplify the denominator:
$ \frac{2\sin \theta\cos^2 \theta}{\cos^2 \theta(1 + \cos \theta)(1 - \cos \theta)} $
Cancel the $ \cos^2 \theta $:
$ \frac{2\sin \theta}{ (1 + \cos \theta)(1 - \cos \theta)} $
Expand the denominator:
$ \frac{2\sin \theta}{1 - \cos^2 \theta} $
Just an identity
$ \frac{2\sin \theta}{\sin^2 \theta} $
There we go
$ 2 \csc \theta $
Solution;
$ \frac{\sec \theta}{\csc \theta - \cot \theta} - \frac{\sec \theta}{\csc \theta + \cot \theta} $
Multiply $ \sin \theta \cos \theta $ above and below:
$ \frac{\sin \theta}{\cos \theta - \cos^2 \theta} - \frac{\sin \theta}{\cos \theta + \cos^2 \theta} $
Make common denominator gives:
$ \frac{\sin \theta(\cos \theta + \cos^2 \theta)}{(\cos \theta + \cos^2 \theta)(\cos \theta - \cos^2 \theta)} - \frac{\sin \theta(\cos \theta - \cos^2 \theta)}{(\cos \theta + \cos^2 \theta)(\cos \theta - \cos^2 \theta)} $
Now do the subtraction
$ \frac{\sin \theta(\cos \theta + \cos^2 \theta)-\sin \theta(\cos \theta - \cos^2 \theta)}{(\cos \theta + \cos^2 \theta)(\cos \theta - \cos^2 \theta)} $
Simplify the numerator by just expand and subtract:
$ \frac{2\sin \theta\cos^2 \theta}{(\cos \theta + \cos^2 \theta)(\cos \theta - \cos^2 \theta)} $
Simplify the denominator:
$ \frac{2\sin \theta\cos^2 \theta}{\cos^2 \theta(1 + \cos \theta)(1 - \cos \theta)} $
Cancel the $ \cos^2 \theta $:
$ \frac{2\sin \theta}{ (1 + \cos \theta)(1 - \cos \theta)} $
Expand the denominator:
$ \frac{2\sin \theta}{1 - \cos^2 \theta} $
Just an identity
$ \frac{2\sin \theta}{\sin^2 \theta} $
There we go
$ 2 \csc \theta $
Sunday, January 10, 2016
Inverse
Problem:
Solution:
Part (a)
$ f^{-1}(x) = \sqrt{15 + x} $
Part (b)
Part (c)
The domain of $ f $ = Range of $ f^{-1} $ = $ [0, +\infty) $.
The range of $ f $ = domain of $ f^{-1} $ = $ [-15, +\infty) $.
Solution:
Part (a)
$ f^{-1}(x) = \sqrt{15 + x} $
Part (b)
The domain of $ f $ = Range of $ f^{-1} $ = $ [0, +\infty) $.
The range of $ f $ = domain of $ f^{-1} $ = $ [-15, +\infty) $.
Subscribe to:
Posts (Atom)