Question

4 cos mũ 2 alpha - 3 sin mũ 2 alpha biết cos anpha = 4/7

Answer 1

Ta có: \(\sin^2\alpha+\cos^2\alpha=1\)

\(\Leftrightarrow\sin^2=1-\dfrac{16}{49}=\dfrac{33}{49}\)

Ta có: \(4\cdot\cos^2\alpha-3\cdot\sin^2\alpha\)

\(=4\cdot\dfrac{16}{49}-3\cdot\dfrac{33}{49}\)

\(=\dfrac{64-99}{49}=-\dfrac{5}{7}\)

Answer 2

`cos^2α=16/49`

`sin^2α+cos^2α=1`

`<=>sin^2α+(4/7)^2=1`

`<=>sin^2α=33/49`

`4cos^2α-3sin^2α=4. 16/49 - 3. 33/49 = -5/7`