Matematică, întrebare adresată de baros13, 8 ani în urmă

6) Folosind formulele de calcul trigonometric, calculati sin 75°, cos 75°, tg 75°, ctg 75° sin 15°, cos 15⁰°, tg 15⁰°, ctg 15⁰°​

Răspunsuri la întrebare

Răspuns de axel0
3

\sin \left(75^{\circ \:}\right) =

\sin \left(x\right)=\cos \left(90^{\circ \:}-x\right)

=\cos \left(90^{\circ \:}-75^{\circ \:}\right)

\cos \left(15^{\circ \:}\right) =\frac{\sqrt{2+\sqrt{3}}}{2}

\cos \left(75^{\circ \:}\right) =

\cos \left(x\right)=\sin \left(90^{\circ \:}-x\right)

=\sin \left(90^{\circ \:}-75^{\circ \:}\right)

\sin \left(15^{\circ \:}\right) =\frac{\sqrt{2-\sqrt{3}}}{2}

\cot \left(75^{\circ \:}\right) =

\cot \left(x\right)=\frac{\cos \left(x\right)}{\sin \left(x\right)}

=\frac{\cos \left(75^{\circ \:}\right)}{\sin \left(75^{\circ \:}\right)} =\frac{\frac{\sqrt{6}-\sqrt{2}}{4}}{\frac{\sqrt{2}+\sqrt{6}}{4}} =2-\sqrt{3}

\sin \left(15^{\circ \:}\right) =

=\sin \left(\frac{30^{\circ \:}}{2}\right)

\sin \left(\frac{x}{2}\right)=\sqrt{\frac{1-\cos \left(x\right)}{2}}

=\sqrt{\frac{1-\cos \left(30^{\circ \:}\right)}{2}}

=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}} =\frac{\sqrt{2-\sqrt{3}}}{2}

\cos \left(15^{\circ \:}\right)=

=\cos \left(\frac{30^{\circ \:}}{2}\right)

\cos \left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos \left(x\right)}{2}}

=\sqrt{\frac{1+\cos \left(30^{\circ \:}\right)}{2}}

=\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}  =\frac{\sqrt{2+\sqrt{3}}}{2}

\tan \left(15^{\circ \:}\right) =

=\sqrt{7-4\sqrt{3}}

=2-\sqrt{3}

\cot \left(15^{\circ \:}\right) =

=\sqrt{7+4\sqrt{3}}

=\sqrt{3}+2

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