Prove (1+sinx)(1-sinx)=cos^{2}x. en. Related Symbolab blog posts. My Notebook, the Symbolab way. Math notebooks have been around for hundreds of years. You write down

Precalculus. Solve for ? cos (x)^2=1/4. cos2 (x) = 1 4 cos 2 ( x) = 1 4. Take the specified root of both sides of the equation to eliminate the exponent on the left side. cos(x) = ±√1 4 cos ( x) = ± 1 4. Simplify ±√1 4 ± 1 4. Tap for more steps cos(x) = ±1 2 cos ( x) = ± 1 2.

So for this question you can use either the product rule or the quotient rule and I'll run through them the quotient rule:The quotient rule says that if you have h(x)=f(x)/g(x)Then h'(x) = (f'(x)g(x)-f(x)g'(x))/(g(x))^2So using f(x)=cos(2x) and g(x)=x^1/2then f'(x)=-2sin(2x) and g'(x)=1/2x^-1/2Plugging this into our formula gives ush(x) = (-2x^1/2sin(2x)-1/2x^-1/2cos(2x))/xAlways remember to simplify afterwards which gives us(-2x^1/2sin(2x)-1/2x^-1/2cos(2x))/xSecond the product rule:What the product rule says is that ifh(x) = f(x)g(x)then h'(x) = f(x)g'(x) + f'(x)g(x)So if we say that h(x) = cos(2x)/x^1/2Then we can say that f(x) = cos(2x) and g(x) = x^-1/2Using the product rule we have:f(x) = cos(2x) f'(x) = -2sin(2x)g(x) = x^-1/2 g'(x) = 1/2x^-3/2So lastly we know that h(x) = f(x)g'(x) + f'(x)g(x)So using what we've found out we can say that h(x) = (cos(2x))/(2x^3/2)-(2sin(2x))/x^1/2Once again simplifying gives us(-2x^1/2sin(2x)-1/2x^-1/2cos(2x))/xNeed help with Maths?One to one online tuition can be a great way to brush up on your Maths a Free Meeting with one of our hand picked tutors from the UK’s top universitiesFind a tutor

As you know there are these trigonometric formulas like Sin 2x, Cos 2x, Tan 2x which are known as double angle formulae for they have double angles in them. To get a good understanding of this topic, Let’s go through the practice examples provided. Cos 2 A = Cos2A – Sin2A = 2Cos2A – 1 = 1 – 2sin2A Introduction to Cos 2 Theta formula Let’s have a look at trigonometric formulae known as the double angle formulae. They are said to be so as it involves double angles trigonometric functions, Cos 2x. Deriving Double Angle Formulae for Cos 2t Let’s start by considering the addition formula. Cos(A + B) = Cos A cos B – Sin A sin B Let’s equate B to A, A = B And then, the first of these formulae becomes: Cos(t + t) = Cos t cos t – Sin t sin t so that Cos 2t = Cos2t – Sin2t And this is how we get second double-angle formula, which is so called because you are doubling the angle (as in 2A). Practice Example for Cos 2: Solve the equation cos 2a = sin a, for – Î \(\begin{array}{l}\leq\end{array} \) a< Î Solution: Let’s use the double angle formula cos 2a = 1 − 2 sin2 a It becomes 1 − 2 sin2 a = sin a 2 sin2 a + sin a − 1=0, Let’s factorise this quadratic equation with variable sinx (2 sin a − 1)(sin a + 1) = 0 2 sin a − 1 = 0 or sin a + 1 = 0 sin a = 1/2 or sin a = −1 To check other mathematical formulas and examples, visit BYJU’S.

cos 2x 1 2