few_shots: 
Example 1:
Concept/skill: 
Prove the addition and subtraction formulas for sine, cosine, and tangent and use them to solve problems.

Problem/activity:
Find the general solution for:
$${\mathrm{cos}\left({\pi\over2}+x\right)=\mathrm{tan}{\pi\over4}}$$


Answer: 
Yes. An initial step when solving for x in this problem requires the use of the addition formula for cosine on the left side of the equation.

Example 2:
Concept/skill: 
Solve quadratic equations in one variable.

Problem/activity:
Task
Solve the quadratic equation
$$ x^2 = (2x - 9)^2 $$
using as many different methods as possible.


Answer: 
Yes. This task shows one quadratic equation with one variable, x.

Example 3:
Concept/skill: 
Construct a tangent line from a point outside a given circle to the circle.

Problem/activity:
Task
Suppose $A$ is an angle with vertex $P$, as pictured below:
[IMG]
Draw a circle with center $P$ and with  radius $r \gt 0$.  Explain why the circle
meets each ray of angle $A$ in a single point.  Label these points $Q$ and $R$ respectively.
Draw circles with centers $Q$ and $R$ respectively and radius $r$.  These circles meet at $P$ and a second point to be labelled $B$.
Show that ray $\overrightarrow{PB}$  bisects angle $P$.


Answer: 
No. This task does not involve drawing lines that are tangent to circles. The only line that may be drawn is the ray between points P and B.

Now, determine whether the following problem or activity can enable students to understand the concepts or skills in the provided description.

shot_info: fl_problem_001894_pos_0

HS_0

im_task_000870_pos_0

HS_1

im_task_001160_neg_0

HS_2
standard_description: Prove theorems about parallelograms. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals.
problem_activity: Task
Below is an isosceles triangle $ABC$ with $|AB| = |AC|$:
[IMG]
Three students propose different arguments for why $m(\angle B) = m(\angle C)$.
Ravi says
If I draw the bisector of $\angle A$ then this is a line of symmetry for $\triangle ABC$ and so $m(\angle B) = m(\angle C)$.
Brittney says
If $M$ is the midpoint of $\overline{BC}$ then $\triangle ABM$ is congruent to $\triangle ACM$ and so $\angle B$ and $\angle C$ are congruent.
Courtney says
If $P$ is a point on $\overline{BC}$ such that $\overleftrightarrow{AP}$ is perpendicular to $\overline{BC}$ then $\triangle ABP$ is congruent to $\triangle ACP$ and so $\angle B$ and $\angle C$ are congruent.
Fill in the details in each argument to show why $m(\angle B) = m(\angle C)$. Can you find another different argument showing that $m(\angle B) = m(\angle C)$?


aligns: no