somerandomperson
New Member
- Joined
- Feb 8, 2019
- Messages
- 16
- Gender
- Male
- HSC
- 2021
I don't get it, sned answrs
-
Finished your HSC? Check out our University Discussion, Non-School forums or help out next year's HSC students! -
YOU can help the next generation of students in the community! Share your trial papers and notes on our Notes & Resources page
Inverse Trig and Identities Questions (1 Viewer)
- Thread starter somerandomperson
- Start date
Inverse Trig questions:
a) easier to follow if you sketch a right-angled triangle, with acute angles A and B so that A + B = 90 deg. Let the sides be x, 1 (hypotenuse) and sqrt(1-x^2) so that sin A = x = cos B and sin B = cos A = sqrt(1-x^2)
 = sin(A - B) = sinAcosB - cosAsinB\\ \\ = x \cdot x - \sqrt{1-x^2} \cdot \sqrt{1-x^2} = x^2 - (1 - x^2) = 2x^2 - 1)
b) too tedious to type out.
Basically use fact that x and
are complementary. You can sketch a right-angled triangle with the 2 acute angles x and , with sin A = x..
a) easier to follow if you sketch a right-angled triangle, with acute angles A and B so that A + B = 90 deg. Let the sides be x, 1 (hypotenuse) and sqrt(1-x^2) so that sin A = x = cos B and sin B = cos A = sqrt(1-x^2)
b) too tedious to type out.
Basically use fact that x and
Last edited:
ii) The line through 
has equation  )
. (can you see why?)
Solve this simultaneously with the equation for the unit circle to obtain a quadratic in
:
x^2+(2t^2)x+(t^2-1)=0)
This equation tells us where the line intersects the circle.
You could just throw the quadratic formula at this, but notice that one of the roots of this quadratic must be
(because clearly the line passes through 
).
The product of roots is
,
and so the other root must be
.
iii) We know
, and circle geometry tells us that 
. So,
\right) = 1=\frac{2t}{1-t^2} )
and now you can easily solve for
.
a) = \sin \arcsin x \cos \arccos x - \cos \arcsin x \sin \arccos x)
Because it is given that
are acute, you can use the right triangle method.
Draw a right triangle with hypotenuse
and sides 
- then one of its angles is 
and the other is 
.
From this we see that
.
So we can show that
 \equiv x^2 - (1-x^2) \equiv 2x^2 -1 )
... for
.
b) Similarly to a) you can draw another right triangle with sides
, but here's another way to do it.
) = \frac{-\sin x}{\sqrt{1-\cos^2x}} = -\frac{\sin x}{|\sin x|} = -1)
(as
for acute angles)
Similarly,
) = -1 )
.
Therefore,
for some constant 
.
Set
to show that 
.
(With this method you can also show with little extra work that both expressions are identically equal to
for acute 
)
Solve this simultaneously with the equation for the unit circle to obtain a quadratic in
This equation tells us where the line intersects the circle.
You could just throw the quadratic formula at this, but notice that one of the roots of this quadratic must be
The product of roots is
and so the other root must be
iii) We know
and now you can easily solve for
a)
Because it is given that
Draw a right triangle with hypotenuse
From this we see that
So we can show that
... for
b) Similarly to a) you can draw another right triangle with sides
(as
Similarly,
Therefore,
Set
(With this method you can also show with little extra work that both expressions are identically equal to