Question

Hi, can someone please help me answer this question?
Determine the exact value of each expression: cos (sin inverse (-square root of 3/2))

Answer 1

@darkknight

Answer 2

\[\cos (\sin^{-1}( -\sqrt{3}/2))\]

Answer 3

this is the problem there are no diagrams

Answer 4

@jhonyy9 ?

Answer 5

ok so we need get the invers of sinus function what is arcsin

y = sin(x) => x = sin(y) => y = arcsin(x)

\[\cos(\sin^{-1}(-\frac{ \sqrt{3} }{ 2 }))\]

We want the value of arcsin(−sqrt3/2) in terms of π.

Let arcsin(−sqrt3/2)=y

⇒siny=−sqrt3/2⇒−siny=sqrt3/2

sin(−y)=−siny⇒sin(−y)=sqrt3/2=sin(π/3)

⇒−y=π/3⇒y=−π/3

⇒arcsin(−sqrt3/2)=−π/3

sin(π/3)=sqrt3/2

sin(π+π/3)=−sin(π/3)=−sqrt3/2

⇒sin(π+π/3)=sin(4π/3)=−sqrt3/2

⇒arcsin(−sqrt3/2)=4π/3

⇒arcsin(−sqrt3/2)=−π/3 or 4π/3

and now we have again \[\cos (\frac{ -\pi }{ 3 }) = - \cos(\frac{ \pi }{ 3 }) = - \frac{ 1 }{ 2 }\]

Answer 6

@supie

Answer 7

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jhonyy9
ok so we need get the invers of sinus function what is arcsin

y = sin(x) => x = sin(y) => y = arcsin(x)

\[\cos(\sin^{-1}(-\frac{ \sqrt{3} }{ 2 }))\]

We want the value of arcsin(−sqrt3/2) in terms of π.

Let arcsin(−sqrt3/2)=y

⇒siny=−sqrt3/2⇒−siny=sqrt3/2

sin(−y)=−siny⇒sin(−y)=sqrt3/2=sin(π/3)

⇒−y=π/3⇒y=−π/3

⇒arcsin(−sqrt3/2)=−π/3

sin(π/3)=sqrt3/2

sin(π+π/3)=−sin(π/3)=−sqrt3/2

⇒sin(π+π/3)=sin(4π/3)=−sqrt3/2

⇒arcsin(−sqrt3/2)=4π/3

⇒arcsin(−sqrt3/2)=−π/3 or 4π/3

and now we have again \[\cos (\frac{ -\pi }{ 3 }) = - \cos(\frac{ \pi }{ 3 }) = - \frac{ 1 }{ 2 }\]
\(\color{#0cbb34}{\text{End of Quote}}\)
@XioGonz

Answer 8

@Vocaloid pls do you agree this ? ty.