Use $$ \cos (A-B)=\cos A \cdot \cos B+\sin A \cdot \sin B $$ to prove $$ \sin (A+B $$

Question

UpStudy AI · Accepted Answer

To prove the identity for $$ \sin(A + B) $$ using the cosine difference formula $$ \cos(A - B) = \cos A \cdot \cos B + \sin A \cdot \sin B $$, we can start by expressing $$ \sin(A + B) $$ in terms of cosine.

We know that:

$$
\sin(A + B) = \sin A \cdot \cos B + \cos A \cdot \sin B
$$

To derive this, we can use the cosine of the complementary angle. Recall that:

$$
\sin(A + B) = \cos\left(\frac{\pi}{2} - (A + B)\right)
$$

Using the cosine difference formula, we can express this as:

$$
\sin(A + B) = \cos\left(\frac{\pi}{2} - A - B\right) = \cos\left(\left(\frac{\pi}{2} - A\right) + B\right)
$$

Now, applying the cosine addition formula:

$$
\cos(X + Y) = \cos X \cdot \cos Y - \sin X \cdot \sin Y
$$

where $$ X = \frac{\pi}{2} - A $$ and $$ Y = B $$:

$$
\cos\left(\frac{\pi}{2} - A + B\right) = \cos\left(\frac{\pi}{2} - A\right) \cdot \cos B - \sin\left(\frac{\pi}{2} - A\right) \cdot \sin B
$$

Using the values of cosine and sine for $$ \frac{\pi}{2} - A $$:

$$
\cos\left(\frac{\pi}{2} - A\right) = \sin A \quad \text{and} \quad \sin\left(\frac{\pi}{2} - A\right) = \cos A
$$

Substituting these into the equation gives:

$$
\sin(A + B) = \sin A \cdot \cos B + \cos A \cdot \sin B
$$

Thus, we have derived the identity:

$$
\sin(A + B) = \sin A \cdot \cos B + \cos A \cdot \sin B
$$

This completes the proof.
To prove $$ \sin(A + B) = \sin A \cdot \cos B + \cos A \cdot \sin B $$, use the cosine difference identity and the complementary angle relationship. By expressing $$ \sin(A + B) $$ as $$ \cos\left(\frac{\pi}{2} - (A + B)\right) $$ and applying the cosine addition formula, we derive the identity. simplify answer: $$ \sin(A + B) = \sin A \cdot \cos B + \cos A \cdot \sin B $$.

Use \( \cos (A-B)=\cos A \cdot \cos B+\sin A \cdot \sin B \) to prove \( \sin (A+B \)

Upstudy AI Solution

Answer

Solution

The Deep Dive

Related Questions

Latest Trigonometry Questions