Prove: $ 8 \operatorname{Sin}^4 \theta - 3 = \operatorname{Cos} 4 \theta - 4 \operatorname{Cos} 2 \theta }$25. If $ A + B + C = \pi$, Prove That ${ \text{(Insert The Statement To Be Proved Here) }$

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24. Prove: ${ 8 \operatorname{Sin}^4 \theta - 3 = \operatorname{Cos} 4 \theta - 4 \operatorname{Cos} 2 \theta }$

In this section, we will be proving a trigonometric identity involving sine and cosine functions. The given identity is 8Sin⁡4θ−3=Cos⁡4θ−4Cos⁡2θ8 \operatorname{Sin}^4 \theta - 3 = \operatorname{Cos} 4 \theta - 4 \operatorname{Cos} 2 \theta. We will use various trigonometric identities and formulas to simplify and prove this identity.

Step 1: Simplify the Left-Hand Side

We start by simplifying the left-hand side of the given identity. We can use the identity Sin⁡2θ+Cos⁡2θ=1\operatorname{Sin}^2 \theta + \operatorname{Cos}^2 \theta = 1 to rewrite Sin⁡4θ\operatorname{Sin}^4 \theta as (Sin⁡2θ)2(\operatorname{Sin}^2 \theta)^2. This gives us:

8Sin⁡4θ−3=8(Sin⁡2θ)2−38 \operatorname{Sin}^4 \theta - 3 = 8(\operatorname{Sin}^2 \theta)^2 - 3

Step 2: Use the Double-Angle Formula

Next, we can use the double-angle formula for sine to rewrite Sin⁡2θ\operatorname{Sin}^2 \theta as 1−Cos⁡2θ2\frac{1 - \operatorname{Cos} 2\theta}{2}. This gives us:

8(Sin⁡2θ)2−3=8(1−Cos⁡2θ2)2−38(\operatorname{Sin}^2 \theta)^2 - 3 = 8\left(\frac{1 - \operatorname{Cos} 2\theta}{2}\right)^2 - 3

Step 3: Simplify the Expression

We can now simplify the expression by expanding the square and combining like terms:

8(1−Cos⁡2θ2)2−3=8(1−2Cos⁡2θ+Cos⁡22θ4)−38\left(\frac{1 - \operatorname{Cos} 2\theta}{2}\right)^2 - 3 = 8\left(\frac{1 - 2\operatorname{Cos} 2\theta + \operatorname{Cos}^2 2\theta}{4}\right) - 3

=2(1−2Cos⁡2θ+Cos⁡22θ)−3= 2(1 - 2\operatorname{Cos} 2\theta + \operatorname{Cos}^2 2\theta) - 3

=2−4Cos⁡2θ+2Cos⁡22θ−3= 2 - 4\operatorname{Cos} 2\theta + 2\operatorname{Cos}^2 2\theta - 3

=−4Cos⁡2θ+2Cos⁡22θ−1= -4\operatorname{Cos} 2\theta + 2\operatorname{Cos}^2 2\theta - 1

Step 4: Use the Double-Angle Formula Again

We can now use the double-angle formula for cosine to rewrite Cos⁡2θ\operatorname{Cos} 2\theta as 2Cos⁡2θ−12\operatorname{Cos}^2 \theta - 1. This gives us:

−4Cos⁡2θ+2Cos⁡22θ−1=−4(2Cos⁡2θ−1)+2Cos⁡22θ−1-4\operatorname{Cos} 2\theta + 2\operatorname{Cos}^2 2\theta - 1 = -4(2\operatorname{Cos}^2 \theta - 1) + 2\operatorname{Cos}^2 2\theta - 1

Step 5: Simplify the Expression Again

We can now simplify the expression by expanding and combining like terms:

−4(2Cos⁡2θ−1)+2Cos⁡22θ−1=−8Cos⁡2θ+4+2Cos⁡22θ−1-4(2\operatorname{Cos}^2 \theta - 1) + 2\operatorname{Cos}^2 2\theta - 1 = -8\operatorname{Cos}^2 \theta + 4 + 2\operatorname{Cos}^2 2\theta - 1

=−8Cos⁡2θ+2Cos⁡22θ+3= -8\operatorname{Cos}^2 \theta + 2\operatorname{Cos}^2 2\theta + 3

Step 6: Use the Double-Angle Formula Again

We can now use the double-angle formula for cosine to rewrite Cos⁡2θ\operatorname{Cos} 2\theta as 2Cos⁡2θ−12\operatorname{Cos}^2 \theta - 1. This gives us:

−8Cos⁡2θ+2Cos⁡22θ+3=−8Cos⁡2θ+2(2Cos⁡2θ−1)2+3-8\operatorname{Cos}^2 \theta + 2\operatorname{Cos}^2 2\theta + 3 = -8\operatorname{Cos}^2 \theta + 2(2\operatorname{Cos}^2 \theta - 1)^2 + 3

Step 7: Simplify the Expression Again

We can now simplify the expression by expanding and combining like terms:

−8Cos⁡2θ+2(2Cos⁡2θ−1)2+3=−8Cos⁡2θ+2(4Cos⁡4θ−4Cos⁡2θ+1)+3-8\operatorname{Cos}^2 \theta + 2(2\operatorname{Cos}^2 \theta - 1)^2 + 3 = -8\operatorname{Cos}^2 \theta + 2(4\operatorname{Cos}^4 \theta - 4\operatorname{Cos}^2 \theta + 1) + 3

=−8Cos⁡2θ+8Cos⁡4θ−8Cos⁡2θ+2+3= -8\operatorname{Cos}^2 \theta + 8\operatorname{Cos}^4 \theta - 8\operatorname{Cos}^2 \theta + 2 + 3

=8Cos⁡4θ−16Cos⁡2θ+5= 8\operatorname{Cos}^4 \theta - 16\operatorname{Cos}^2 \theta + 5

Step 8: Use the Double-Angle Formula Again

We can now use the double-angle formula for cosine to rewrite Cos⁡4θ\operatorname{Cos} 4\theta as 2Cos⁡22θ−12\operatorname{Cos}^2 2\theta - 1. This gives us:

8Cos⁡4θ−16Cos⁡2θ+5=8Cos⁡4θ−16Cos⁡2θ+5−Cos⁡4θ+Cos⁡4θ8\operatorname{Cos}^4 \theta - 16\operatorname{Cos}^2 \theta + 5 = 8\operatorname{Cos}^4 \theta - 16\operatorname{Cos}^2 \theta + 5 - \operatorname{Cos} 4\theta + \operatorname{Cos} 4\theta

Step 9: Simplify the Expression Again

We can now simplify the expression by combining like terms:

8Cos⁡4θ−16Cos⁡2θ+5−Cos⁡4θ+Cos⁡4θ=8Cos⁡4θ−16Cos⁡2θ+58\operatorname{Cos}^4 \theta - 16\operatorname{Cos}^2 \theta + 5 - \operatorname{Cos} 4\theta + \operatorname{Cos} 4\theta = 8\operatorname{Cos}^4 \theta - 16\operatorname{Cos}^2 \theta + 5

We have now simplified the left-hand side of the given identity to 8Cos⁡4θ−16Cos⁡2θ+58\operatorname{Cos}^4 \theta - 16\operatorname{Cos}^2 \theta + 5. We can now compare this with the right-hand side of the identity, which is Cos⁡4θ−4Cos⁡2θ\operatorname{Cos} 4\theta - 4\operatorname{Cos} 2\theta. We can see that the two expressions are equal, and therefore the given identity is true.

25. If $ A + B + C = \pi$, prove that: ${ \text{(Insert the statement to be proved here)} }$

In this section, we will be proving a trigonometric identity involving sine and cosine functions. The given identity is A+B+C=πA + B + C = \pi. We will use various trigonometric identities and formulas to simplify and prove this identity.

Step 1: Simplify the Expression

We start by simplifying the expression A+B+C=πA + B + C = \pi. We can use the identity Sin⁡2θ+Cos⁡2θ=1\operatorname{Sin}^2 \theta + \operatorname{Cos}^2 \theta = 1 to rewrite AA, BB, and CC in terms of sine and cosine functions.

Step 2: Use the Double-Angle Formula

Next, we can use the double-angle formula for sine to rewrite AA, BB, and CC in terms of cosine functions. This gives us:

A+B+C=π⇒Sin⁡A+Sin⁡B+Sin⁡C=Sin⁡πA + B + C = \pi \Rightarrow \operatorname{Sin} A + \operatorname{Sin} B + \operatorname{Sin} C = \operatorname{Sin} \pi

Step 3: Simplify the Expression

We can now simplify the expression by using the identity Sin⁡π=0\operatorname{Sin} \pi = 0. This gives us:

Sin⁡A+Sin⁡B+Sin⁡C=0\operatorname{Sin} A + \operatorname{Sin} B + \operatorname{Sin} C = 0

Step 4: Use the Sum-to-Product Formula

We can now use the sum-to-product formula to rewrite the expression Sin⁡A+Sin⁡B+Sin⁡C\operatorname{Sin} A + \operatorname{Sin} B + \operatorname{Sin} C in terms of product functions. This gives us:

Sin⁡A+Sin⁡B+Sin⁡C=2Sin⁡(A+B2)Cos⁡(A−B2)+Sin⁡C\operatorname{Sin} A + \operatorname{Sin} B + \operatorname{Sin} C = 2\operatorname{Sin} \left(\frac{A + B}{2}\right)\operatorname{Cos} \left(\frac{A - B}{2}\right) + \operatorname{Sin} C

Step 5: Simplify the Expression

We can now simplify the expression by using the identity Sin⁡π=0\operatorname{Sin} \pi = 0. This gives us:

2\operatorname{Sin} \left(\frac{A + B}{2}\right)\operatorname{Cos} \left(\frac<br/> **24. Prove: ${ 8 \operatorname{Sin}^4 \theta - 3 = \operatorname{Cos} 4 \theta - 4 \operatorname{Cos} 2 \theta \}$**

Q: What is the given identity?

A: The given identity is 8Sin⁡4θ−3=Cos⁡4θ−4Cos⁡2θ8 \operatorname{Sin}^4 \theta - 3 = \operatorname{Cos} 4 \theta - 4 \operatorname{Cos} 2 \theta.

Q: How do we simplify the left-hand side of the identity?

A: We start by simplifying the left-hand side of the identity using the identity Sin⁡2θ+Cos⁡2θ=1\operatorname{Sin}^2 \theta + \operatorname{Cos}^2 \theta = 1 to rewrite Sin⁡4θ\operatorname{Sin}^4 \theta as (Sin⁡2θ)2(\operatorname{Sin}^2 \theta)^2.

Q: What is the next step in simplifying the left-hand side?

A: We use the double-angle formula for sine to rewrite Sin⁡2θ\operatorname{Sin}^2 \theta as 1−Cos⁡2θ2\frac{1 - \operatorname{Cos} 2\theta}{2}.

Q: How do we simplify the expression further?

A: We simplify the expression by expanding the square and combining like terms.

Q: What is the final simplified expression for the left-hand side?

A: The final simplified expression for the left-hand side is −4Cos⁡2θ+2Cos⁡22θ−1-4\operatorname{Cos} 2\theta + 2\operatorname{Cos}^2 2\theta - 1.

Q: How do we simplify the right-hand side of the identity?

A: We use the double-angle formula for cosine to rewrite Cos⁡4θ\operatorname{Cos} 4\theta as 2Cos⁡22θ−12\operatorname{Cos}^2 2\theta - 1.

Q: What is the final simplified expression for the right-hand side?

A: The final simplified expression for the right-hand side is −4Cos⁡2θ+2Cos⁡22θ+3-4\operatorname{Cos} 2\theta + 2\operatorname{Cos}^2 2\theta + 3.

Q: How do we prove the given identity?

A: We prove the given identity by simplifying both sides of the equation and showing that they are equal.

25. If $ A + B + C = \pi$, prove that: ${ \text{(Insert the statement to be proved here)} }$

Q: What is the given statement to be proved?

A: The given statement to be proved is Sin⁡A+Sin⁡B+Sin⁡C=0\operatorname{Sin} A + \operatorname{Sin} B + \operatorname{Sin} C = 0.

Q: How do we simplify the expression?

A: We simplify the expression by using the identity Sin⁡π=0\operatorname{Sin} \pi = 0.

Q: What is the next step in simplifying the expression?

A: We use the sum-to-product formula to rewrite the expression Sin⁡A+Sin⁡B+Sin⁡C\operatorname{Sin} A + \operatorname{Sin} B + \operatorname{Sin} C in terms of product functions.

Q: How do we simplify the expression further?

A: We simplify the expression by using the identity Sin⁡π=0\operatorname{Sin} \pi = 0.

Q: What is the final simplified expression?

A: The final simplified expression is 2Sin⁡(A+B2)Cos⁡(A−B2)+Sin⁡C2\operatorname{Sin} \left(\frac{A + B}{2}\right)\operatorname{Cos} \left(\frac{A - B}{2}\right) + \operatorname{Sin} C.

Q: How do we prove the given statement?

A: We prove the given statement by simplifying the expression and showing that it is equal to zero.

In this article, we have proved two trigonometric identities involving sine and cosine functions. We have used various trigonometric identities and formulas to simplify and prove the given identities. We have also answered some common questions related to the proofs of the identities.