Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

If tanA=34\tan A = \frac{3}{4}, prove that sinAcosA=1225\sin A \cos A = \frac{12}{25}

Full solution

Q. If tanA=34\tan A = \frac{3}{4}, prove that sinAcosA=1225\sin A \cos A = \frac{12}{25}
  1. Write Trig Identity: We know that tanA=sinAcosA\tan A = \frac{\sin A}{\cos A}. Given tanA=34\tan A = \frac{3}{4}, we can write sinAcosA=34\frac{\sin A}{\cos A} = \frac{3}{4}.
  2. Apply Pythagorean Identity: To find sinA\sin A and cosA\cos A, we will use the Pythagorean identity: sin2A+cos2A=1\sin^2 A + \cos^2 A = 1.
  3. Assume Values for sinA\sin A and cosA\cos A: Let's assume sinA=3k\sin A = \frac{3}{k} and cosA=4k\cos A = \frac{4}{k} for some kk, because tanA=sinAcosA=34\tan A = \frac{\sin A}{\cos A} = \frac{3}{4}. We need to find the value of kk.
  4. Substitute Values in Pythagorean Identity: Using the Pythagorean identity, we substitute sinA\sin A and cosA\cos A with the assumed values: (3k)2+(4k)2=1(\frac{3}{k})^2 + (\frac{4}{k})^2 = 1.
  5. Solve for k: Solving the equation for k, we get 9k2+16k2=1\frac{9}{k^2} + \frac{16}{k^2} = 1.
  6. Find Value of k: Combining the terms, we have (9+16)/k2=1(9 + 16) / k^2 = 1, which simplifies to 25/k2=125/k^2 = 1.
  7. Calculate sinA\sin A and cosA\cos A: Solving for k2k^2, we find k2=25k^2 = 25.
  8. Multiply sinA\sin A and cosA\cos A: Taking the square root of both sides, we get k=5k = 5, since kk is positive (kk represents a length in the context of a right triangle).
  9. Prove Sin A Cos A: Now we can find sinA\sin A and cosA\cos A using the values of kk: sinA=3k=35\sin A = \frac{3}{k} = \frac{3}{5} and cosA=4k=45\cos A = \frac{4}{k} = \frac{4}{5}.
  10. Prove Sin A Cos A: Now we can find sinA\sin A and cosA\cos A using the values of kk: sinA=3k=35\sin A = \frac{3}{k} = \frac{3}{5} and cosA=4k=45\cos A = \frac{4}{k} = \frac{4}{5}.To prove sinAcosA=1225\sin A \cos A = \frac{12}{25}, we multiply the values we found: sinAcosA=(35)(45)\sin A \cdot \cos A = \left(\frac{3}{5}\right) \cdot \left(\frac{4}{5}\right).
  11. Prove Sin A Cos A: Now we can find sinA\sin A and cosA\cos A using the values of kk: sinA=3k=35\sin A = \frac{3}{k} = \frac{3}{5} and cosA=4k=45\cos A = \frac{4}{k} = \frac{4}{5}.To prove sinAcosA=1225\sin A \cos A = \frac{12}{25}, we multiply the values we found: sinAcosA=(35)(45)\sin A \cdot \cos A = \left(\frac{3}{5}\right) \cdot \left(\frac{4}{5}\right).Calculating the product, we get sinAcosA=1225\sin A \cdot \cos A = \frac{12}{25}, which is what we wanted to prove.

More problems from Find a value using two-variable equations