Solve

[courteous]

For all 3 problems the following is true: \(\displaystyle \sin\alpha+\cos\alpha=A\); \(\displaystyle |A|\leq\sqrt2\).

1) \(\displaystyle \sin\alpha*\cos\alpha=\) (got the first one) \(\displaystyle =\frac{(\sin\alpha+\cos\alpha)^2-1}{2}=\frac{A^2-1}{2}\)

2) \(\displaystyle \vline\cos\alpha - \sin\alpha\vline=\) :?:

3) \(\displaystyle \sin^4{\alpha}+\cos^4{\alpha}=\) :?:

 

[Deleted member 4993]

For all 3 problems the following is true: \(\displaystyle \sin\alpha+\cos\alpha=A\); \(\displaystyle |A|\leq\sqrt2\).

1) \(\displaystyle \sin\alpha*\cos\alpha=\) (got the first one) \(\displaystyle =\frac{(\sin\alpha+\cos\alpha)^2-1}{2}=\frac{A^2-1}{2}\)

2) \(\displaystyle \vline\cos\alpha - \sin\alpha\vline=\) :?:

for (2) first find

\(\displaystyle \cos\alpha - \sin\alpha = \pm \sqrt {[ \cos\alpha + \sin\alpha]^2 - 4\cdot \sin\alpha*\cos\alpha}\)

3) \(\displaystyle \sin^4{\alpha}+\cos^4{\alpha}=\) :?:

Looking at hint for (2) - can you derive a hint for this one?

 

[courteous]

For all 3 problems the following is true: \(\displaystyle \sin\alpha+\cos\alpha=A\); \(\displaystyle |A|\leq\sqrt2\).

1) \(\displaystyle \sin\alpha*\cos\alpha=\) (got the first one) \(\displaystyle =\frac{(\sin\alpha+\cos\alpha)^2-1}{2}=\frac{A^2-1}{2}\)

2) \(\displaystyle \vline\cos\alpha - \sin\alpha\vline=\) :?:

for (2) first find

\(\displaystyle \cos\alpha - \sin\alpha = \pm \sqrt {[ \cos\alpha + \sin\alpha]^2 - 4\cdot \sin\alpha*\cos\alpha}\)

3) \(\displaystyle \sin^4{\alpha}+\cos^4{\alpha}=\) :?:

Looking at hint for (2) - can you derive a hint for this one?

3) \(\displaystyle \sin^4{\alpha}+\cos^4{\alpha}=(sin^2\alpha+cos^2\alpha)^2-2\sin^2\alpha\cos^2\alpha\) I thank you.