back to *trigonometry video lessons*
# Compare sin55^{o}, cos55^{o} and tan55^{o} without use a calculator

Question:

Do not use a calculator, compare the values of sin 55^{o}, cos 55^{o} and tan 55^{o}. Writing then in ascending order.

Solution:

The circle is a unit circle, The radius of the circle is one. r = 1.

The vertex of the angle is in origin. The initial side of the angle is the positive x-axis, the angle a = 55^{o}. The terminal side of the angle is in Quadrant I.

The terminal side of the angle intersects the unit circle at point P. Draw a line passes the point P and perpendicular to the positive x-axis, this line intersects the positive x-axis at point A. Point A is a perpendicular point.

Because PA is perpendicular to the positive x-axis, so, the angle PAO is 90^{o}. So, triangle PAO is a right triangle. In right triangle OPA,

By sine definition, sin 55^{o} = opposite side/Hypotenuse = PA/OP = PA/r = PA/1 = PA. PA is sine line and PA > 0.

By cosine definition, cos 55^{o} = adjacent side/Hypotenuse = OA/OP = OA/r = OA /1 = OA. OA is cosine line and OA > 0.

The unit circle intersects the positive x-axis at point B. Draw a line passes the point B and perpendicular to the positive x-axis. This line intersects the terminal side of the angle at
point Q. Because the point B is a perpendicular point, so the triangle OBQ is a right triangle. In right triangle OBQ, tan 55^{o} = (opposite side)/(adjacent side) = QB/OB = QB/r = QB/1 = QB.
QB is the tangent line and QB > 0. Look the figure, QB > PA, so tan 55^{o} > sin 55^{o}

In right triangle OAP, because the angle POA = 55^{o}, so the angle OPA = 90^{o} - 55^{o} = 35^{o}. In triangle OPA, the side opposite 35^{o} is less than
the side opposite 55^{o}. So, the length of OA is less than the length of PA. Because the length of PA is less than the length of QB, so, OA < PA < QB. Because OA is the cosine line,
PA is the sine line and QB is the tangent line, So, cos 55^{o} < sin 55^{o} < tan 55^{o}. Please watch the video for more details.