Solve cosθ90

$$\cos \theta 90$$

Differentiate w.r.t. θ_90

$$-\sin(\theta _{90})$$

Steps Using Definition of a Derivative

For a function f\left(x\right), the derivative is the limit of \frac{f\left(x+h\right)-f\left(x\right)}{h} as h goes to 0, if that limit exists.

$$\frac{\mathrm{d}}{\mathrm{d}\theta _{90}}(\cos(\theta _{90}))=\left(\lim_{h\to 0}\frac{\cos(\theta _{90}+h)-\cos(\theta _{90})}{h}\right)$$

Use the Sum Formula for Cosine.

$$\lim_{h\to 0}\frac{\cos(h+\theta _{90})-\cos(\theta _{90})}{h}$$

Factor out \cos(\theta _{90}).

$$\lim_{h\to 0}\frac{\cos(\theta _{90})\left(\cos(h)-1\right)-\sin(\theta _{90})\sin(h)}{h}$$

Rewrite the limit.

$$\left(\lim_{h\to 0}\cos(\theta _{90})\right)\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)-\left(\lim_{h\to 0}\sin(\theta _{90})\right)\left(\lim_{h\to 0}\frac{\sin(h)}{h}\right)$$

Use the fact that \theta _{90} is a constant when computing limits as h goes to 0.

$$\cos(\theta _{90})\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)-\sin(\theta _{90})\left(\lim_{h\to 0}\frac{\sin(h)}{h}\right)$$

The limit \lim_{\theta _{90}\to 0}\frac{\sin(\theta _{90})}{\theta _{90}} is 1.

$$\cos(\theta _{90})\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)-\sin(\theta _{90})$$

To evaluate the limit \lim_{h\to 0}\frac{\cos(h)-1}{h}, first multiply the numerator and denominator by \cos(h)+1.

$$\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)=\left(\lim_{h\to 0}\frac{\left(\cos(h)-1\right)\left(\cos(h)+1\right)}{h\left(\cos(h)+1\right)}\right)$$

Multiply \cos(h)+1 times \cos(h)-1.

$$\lim_{h\to 0}\frac{\left(\cos(h)\right)^{2}-1}{h\left(\cos(h)+1\right)}$$

Use the Pythagorean Identity.

$$\lim_{h\to 0}-\frac{\left(\sin(h)\right)^{2}}{h\left(\cos(h)+1\right)}$$

Rewrite the limit.

$$\left(\lim_{h\to 0}-\frac{\sin(h)}{h}\right)\left(\lim_{h\to 0}\frac{\sin(h)}{\cos(h)+1}\right)$$

The limit \lim_{\theta _{90}\to 0}\frac{\sin(\theta _{90})}{\theta _{90}} is 1.

$$-\left(\lim_{h\to 0}\frac{\sin(h)}{\cos(h)+1}\right)$$

Use the fact that \frac{\sin(h)}{\cos(h)+1} is continuous at 0.

$$\left(\lim_{h\to 0}\frac{\sin(h)}{\cos(h)+1}\right)=0$$

Substitute the value 0 into the expression \cos(\theta _{90})\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)-\sin(\theta _{90}).

$$-\sin(\theta _{90})$$

Evaluate

$$\cos(\theta _{90})$$