Question

Prove that sinx/1-cosx + 1-cosx/sinx = 2cosex

Answer 1

Answer:

PLEASE MARK ME AS BRAINLISTED

Step-by-step explanation:

Answer 2

Answer:

1

3

How do I prove that (sinx+cosx+1)(sinx+cosx−1)=sin2x ?

Learn while you earn with a Work Integrated Learning Program.

William Mccoy

Answered May 24, 2017

The given trigonometric equation can be proven by using the appropriate algebraic manipulations and Trigonometric identities as follows:

(sin x + cos x + 1)(sin x + cos x ‒ 1) = sin 2x

After performing the indicated multiplication shown on the left and adding the partial products, we have the following product:

sin² x + 2 sin x cos x + sin x ‒ sin x + cos² x + cos x ‒ cos x ‒ 1 = sin 2x

sin² x + 2 sin x cos x + (sin x ‒ sin x) + cos² x + (cos x‒cos x) ‒ 1 = sin 2x

By the Distributive Property of Real Numbers, we have:

sin² x + 2 sin x cos x + (1 ‒ 1)sin x + cos² x + (1 ‒ 1)cos x ‒ 1 = sin 2x

sin² x + 2 sin x cos x + (0)sin x + cos² x + (0)cos x ‒ 1 = sin 2x

sin² x + 2 sin x cos x + 0 + cos² x + 0 ‒ 1 = sin 2x

sin² x + 2 sin x cos x + cos² x ‒ 1 = sin 2x

Now, by the Commutative Property and of Addition, we have:

sin² x + cos² x + 2sin x cos x ‒ 1 = sin 2x

Now, by the Associative Property of Addition, we have:

(sin² x + cos² x) + 2sin x cos x ‒ 1 = sin 2x

Since one of the basic Trigonometric Identities is:

sin² x + cos² x = 1, for any angle x, then substituting, we have:

(1)+ 2 sin x cos x ‒ 1 = sin 2x

2 sin x cos x + 1 ‒ 1 = sin 2x

2 sin x cos x + (1 ‒ 1) = sin 2x

2 sinx cos x + 0 = sin 2x

2 sinx cos x = sin 2x

Now, by one of the Double-Angle Trigonometric Identities, we have:

sin 2x = 2 sin x cos x; therefore, substituting on the right, we finally get:

2 sin x cos x = 2 sin x cos x

We now see that the given trigonometric equation is verified (proven) and is therefore an identity that is true for all permissible values of the angle x.

Are you ready to power your career by maximising your learnings?

Karina Hastings

Answered January 24, 2019

Whenever doing calculations of trigonometric functions it is always helpful to have a good understanding of basic trigonometric properties and formulas. For this particular question we will need to be aware of the Pythagorean Identity for sine and cosine, which is sin2x+cos2x=1 , and the double-angle formula for sine, which is sin2x=2sinx⋅cosx . So now let’s multiply our two polynomials from the left side of the equation together. First, we need to multiply each term in one polynomial by each term in the other polynomial, so we get:

sin2x+sinx⋅cosx−sinx+cosx⋅sinx+cos2x−cosx+sinx+cosx−1

Next, we combine all like terms and simplify:

sin2x+cos2x+2sinx⋅cosx−1

Now we can see how the Pythagorean Id

Mathematics Solutions

Mathematics Concepts

rohits education

Mathematics

Math & Science

For maths & science

Science And Math

world of mathematics

Science n math adventures

Math Club

Tazwar

Answered April 12, 2017

We have: (sin(x)+cos(x)+1)(sin(x)+cos(x)−1)

=((sin(x)+cos(x))+1)((sin(x)+cos(x))−1)

Let’s factor this into a difference of squares:

=(sin(x)+cos(x))2−(1)2

=sin2(x)+2sin(x)cos(x)+cos2(x)−1

One of the Pythagorean identities is cos2(x)+sin2(x)=1 .

Let’s rearrange this as:

⇒cos2(x)=1−sin2(x)

Then, let’s apply this rearranged identity to our proof:

=sin2(x)+2sin(x)cos(x)+(1−sin2(x))−1

=2sin(x)cos(x)

Now, let’s apply the double angle identity for sin(x) ; sin(2x)=2sin(x)cos(x) :

=sin(2x) Q.E.D.

Step-by-step explanation:

please like my answer