\sin(T_ANG
OP S_ANG) = \; ?
ANS_DISPLAYop
We don't know what
T_ANG is exactly,
so we can't directly evaluate this function. We do know
what \sin(T_ANG) is,
though.
To simplify this formula to something we can use, we try
the sine addition/subtraction identity:
\sin(x \pm y)
= \sin x \cdot \cos y \pm \cos x \cdot \sin y
In this case, we have
\qquad \sin(T_ANG
OP S_ANG) =
\qquad\qquad
\sin(T_ANG) \cdot
\cos(S_ANG) OP
\cos(T_ANG) \cdot
\sin(S_ANG)
Now we just need to evaluate each term.
\qquad \sin(T_ANG) =
\dfrac{Opposite}{Hypotenuse} =
\dfrac{OPPOSITE_NAME}
{HYPOTENUSE_NAME} =
TERM1
\qquad \cos(S_ANG) =
TERM2
\qquad \cos(T_ANG) =
\dfrac{Adjacent}{Hypotenuse} =
\dfrac{ADJACENT_NAME}
{HYPOTENUSE_NAME} =
TERM3
\qquad \sin(S_ANG) =
TERM4
Putting it together, we get
\qquad TERM1 \cdot TERM2
OP TERM3 \cdot TERM4
= ANS_DISPLAY
\cos(T_ANG
OP S_ANG) = \; ?
ANS_DISPLAYop
We don't know what
T_ANG is exactly,
so we can't directly evaluate this function. We do know
what \cos(T_ANG) is,
though.
To simplify this formula to something we can use, we try
the cosine addition/subtraction identity:
\cos(x \pm y)
= \cos x \cdot \cos y \mp \sin x \cdot \sin y
In this case, we have
\qquad \cos(T_ANG
OP S_ANG) =
\qquad\qquad
\cos(T_ANG) \cdot
\cos(S_ANG) OP2
\sin(T_ANG) \cdot
\sin(S_ANG)
Now we just need to evaluate each term.
\qquad \cos(T_ANG) =
\dfrac{Adjacent}{Hypotenuse} =
\dfrac{ADJACENT_NAME}
{HYPOTENUSE_NAME} =
TERM1
\qquad \cos(S_ANG) =
TERM2
\qquad \sin(T_ANG) =
\dfrac{Opposite}{Hypotenuse} =
\dfrac{OPPOSITE_NAME}
{HYPOTENUSE_NAME} =
TERM3
\qquad \sin(S_ANG) =
TERM4
Putting it together, we get
\qquad TERM1 \cdot TERM2
OP2 TERM3 \cdot TERM4
= ANS_DISPLAY
\sin(2 \cdot T_ANG) = \; ?
We don't know what
T_ANG is exactly, so we can't
compute 2 \cdot T_ANG to directly
evaluate this function. We do know what \sin(
T_ANG) and \cos(T_ANG)
are, though.
To simplify this formula to something we can use, we try
the sine double-angle identity:
\sin(2x) = 2 \sin (x) \cos (x)
In this case, we have
\qquad \sin(2 \cdot T_ANG) =
2 \sin(T_ANG)
\cos(T_ANG)
(To cut down on the number of identities you have to memorize, you can derive this quickly from the angle addition identity for sine) [ Show how]
Start with the sine angle addition identity:
\qquad \sin(x + y) = \sin(x) \cdot \cos(y)
+ \cos(x) \cdot \sin(y)
Now take the case where x = y:
\qquad \sin(x + y) = \sin(x + x)
= \sin(x) \cdot \cos(x) + \cos(x) \cdot \sin(x)
\qquad \sin(2x) = 2 \sin(x) \cos(x)
Now we just need to evaluate each term.
\qquad \sin(T_ANG) =
\dfrac{Opposite}{Hypotenuse} =
\dfrac{OPPOSITE_NAME}
{HYPOTENUSE_NAME} =
TERM1
\qquad \cos(T_ANG) =
\dfrac{Adjacent}{Hypotenuse} =
\dfrac{ADJACENT_NAME}
{HYPOTENUSE_NAME} =
TERM2
Putting it together, we get
\qquad 2 \cdot TERM1
\cdot TERM2
= ANS_DISPLAY
\cos(2 \cdot T_ANG) = \; ?
We don't know what
T_ANG is exactly, so we can't
compute 2 \cdot T_ANG to directly
evaluate this function. We do know what \sin(
T_ANG) and \cos(T_ANG)
are, though.
To simplify this formula to something we can use, we try
the cosine double-angle identity:
\cos(2x) = \cos^2 (x) - \sin^2 (x)
In this case, we have
\qquad \cos(2 \cdot T_ANG) =
\cos^2(T_ANG) -
\sin^2(T_ANG)
(To cut down on the number of identities you have to memorize, you can derive this quickly from the angle addition identity for cosine) [ Show how]
Start with the cosine angle addition identity:
\qquad \cos(x + y) = \cos(x) \cdot \cos(y)
- \sin(x) \cdot \sin(y)
Now take the case where x = y:
\qquad \cos(x + y) = \cos(x + x)
= \cos(x) \cdot \cos(x) - \sin(x) \cdot \sin(x)
\qquad \cos(2x) = \cos^2(x) - \sin^2(x)
Now we just need to evaluate each term.
\qquad \cos^2(T_ANG) =
\left(\dfrac{Adj}{Hyp}\right)^2 =
\left(\dfrac{ADJACENT_NAME}
{HYPOTENUSE_NAME}\right)^2 =
\left(\dfrac{ADJACENT_VALUE}
{HYPOTENUSE_VALUE}\right)^2 =
TERM1
\qquad \sin^2(T_ANG) =
\left(\dfrac{Opp}{Hyp}\right)^2 =
\left(\dfrac{OPPOSITE_NAME}
{HYPOTENUSE_NAME}\right)^2 =
\left(\dfrac{OPPOSITE_VALUE}
{HYPOTENUSE_VALUE}\right)^2 =
TERM2
Putting it together, we get
\qquad TERM1
- TERM2
= ANS_DISPLAY